What are the basic rules and idioms for operator overloading?
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Note: This question and the original answers are from 2010 and partially outdated. Most of it is still good and helpful, but the original text no longer covers everything there is to know about C++ operator overloading (e.g., <=>, [] can now be multidimensional). This has been fixed by others over the years, but please keep in mind that I do not longer actively maintain this FAQ. — sbi, 2024-02-03

The answers were given in a specific order, but since most users sort answers according to votes, rather than the time they were given, here's an index of the answers in the order in which they make the most sense:

(Note: This is meant to be an entry to Stack Overflow's C++ FAQ. If you want to critique the idea of providing an FAQ in this form, then the posting on meta that started all this would be the place to do that. Answers to that question are monitored in the C++ chatroom, where the FAQ idea started in the first place, so your answer is very likely to get read by those who came up with the idea.)

Kati answered 12/12, 2010 at 12:44 Comment(2)
If we're going to continue with the C++-FAQ tag, this is how entries should be formatted.Helvetii
Oh, and an English translation is also available: the basics and common practiceIshtar
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1204

Common Operators to Overload

Most of the work in overloading operators is boilerplate code. That is little wonder, since operators are merely syntactic sugar. Their actual work could be done by (and often is forwarded to) plain functions. But it is important that you get this boilerplate code right. If you fail, either your operator’s code won’t compile, your users’ code won’t compile, or your users’ code will behave surprisingly.

Assignment Operator

There's a lot to be said about assignment. However, most of it has already been said in GMan's famous Copy-And-Swap FAQ, so I'll skip most of it here, only listing the perfect assignment operator for reference:

X& X::operator=(X rhs)
{
  swap(rhs);
  return *this;
}

Stream Insertion and Extraction

Disclaimer
For overloading << and >> as bitwise shift operators, skip to the section Binary Arithmetic Operators.

The bitwise shift operators << and >>, although still used in hardware interfacing for the bit-manipulation functions they inherit from C, have become more prevalent as overloaded stream input and output operators in most applications.

The stream operators, among the most commonly overloaded operators, are binary infix operators for which the syntax does not specify any restriction on whether they should be members or non-members. However, their left operands are streams from the standard library, and you cannot add member functions to those1, so you need to implement these operators for your own types as non-member functions2. The canonical forms of the two are these:

std::ostream& operator<<(std::ostream& os, const T& obj)
{
  // Write obj to stream
  return os;
}

std::istream& operator>>(std::istream& is, T& obj)
{
  // Read obj from stream
  if( /* no valid object of T found in stream */ )
    is.setstate(std::ios::failbit);
  return is;
}

When implementing operator>>, manually setting the stream’s state is only necessary when the reading itself succeeded, but the result is not what would be expected.

1 Note that some of the << overloads of the standard library are implemented as member functions, and some as free functions. Only the locale-dependent functions are member functions, such as operator<<(long).

2 According to the rules of thumb, the insertion/extraction operators should be member functions because they modify the left operand. However, we cannot follow the rules of thumb here.

Function Call Operator

The function call operator, used to create function objects, also known as functors, must be defined as a member function, so it always has the implicit this argument of member functions. Other than this, it can be overloaded to take any number of additional arguments, including zero.

Here's an example of the syntax:

struct X {
    // Overloaded call operator
    int operator()(const std::string& y) {
        return /* ... */;
    }
};

Usage:

X f;
int a = f("hello");

Throughout the C++ standard library, function objects are always copied. Your own function objects should therefore be cheap to copy. If a function object absolutely needs to use data which is expensive to copy, it is better to store that data elsewhere and have the function object refer to it.

Comparison Operators

This section has been moved elsewhere
See this FAQ answer for overloading the binary infix ==, !=, <, >, <=, and >= operators, as well as the <=> three-way comparison, aka. "spaceship operator" in C++20. There is so much to say about comparison operators that it would exceed the scope of this answer.

In the most simple case, you can overload all comparison comparison operators by defaulting <=> in C++20:

#include <compare>

struct X {
  // defines ==, !=, <, >, <=, >=, <=>
  friend auto operator<=>(const X&, const X&) = default;
};

If you can't do this, continue to the linked answer.

Logical Operators

The unary prefix negation ! should be implemented as a member function. It is usually not a good idea to overload it because of how rare and surprising it is.

struct X {
  X operator!() const { return /* ... */; }
};

The remaining binary logical operators (||, &&) should be implemented as free functions. However, it is very unlikely that you would find a reasonable use case for these1.

X operator&&(const X& lhs, const X& rhs) { return /* ... */; }
X operator||(const X& lhs, const X& rhs) { return /* ... */; }

1 It should be noted that the built-in version of || and && use shortcut semantics. While the user defined ones (because they are syntactic sugar for method calls) do not use shortcut semantics. User will expect these operators to have shortcut semantics, and their code may depend on it, Therefore it is highly advised NEVER to define them.

Arithmetic Operators

Unary Arithmetic Operators

The unary increment and decrement operators come in both prefix and postfix flavor. To tell one from the other, the postfix variants take an additional dummy int argument. If you overload increment or decrement, be sure to always implement both prefix and postfix versions.

Here is the canonical implementation of increment, decrement follows the same rules:

struct X {
  X& operator++()
  {
    // Do actual increment
    return *this;
  }
  X operator++(int)
  {
    X tmp(*this);
    operator++();
    return tmp;
  }
};

Note that the postfix variant is implemented in terms of prefix. Also note that postfix does an extra copy.1

Overloading unary minus and plus is not very common and probably best avoided. If needed, they should probably be overloaded as member functions.

1 Also note that the postfix variant does more work and is therefore less efficient to use than the prefix variant. This is a good reason to generally prefer prefix increment over postfix increment. While compilers can usually optimize away the additional work of postfix increment for built-in types, they might not be able to do the same for user-defined types (which could be something as innocently looking as a list iterator). Once you got used to do i++, it becomes very hard to remember to do ++i instead when i is not of a built-in type (plus you'd have to change code when changing a type), so it is better to make a habit of always using prefix increment, unless postfix is explicitly needed.

Binary Arithmetic Operators

For the binary arithmetic operators, do not forget to obey the third basic rule operator overloading: If you provide +, also provide +=, if you provide -, do not omit -=, etc. Andrew Koenig is said to have been the first to observe that the compound assignment operators can be used as a base for their non-compound counterparts. That is, operator + is implemented in terms of +=, - is implemented in terms of -=, etc.

According to our rules of thumb, + and its companions should be non-members, while their compound assignment counterparts (+=, etc.), changing their left argument, should be a member. Here is the exemplary code for += and +; the other binary arithmetic operators should be implemented in the same way:

struct X {
  X& operator+=(const X& rhs)
  {
    // actual addition of rhs to *this
    return *this;
  }
};

inline X operator+(const X& lhs, const X& rhs)
{
  X result = lhs;
  result += rhs;
  return result;
}

operator+= returns its result per reference, while operator+ returns a copy of its result. Of course, returning a reference is usually more efficient than returning a copy, but in the case of operator+, there is no way around the copying. When you write a + b, you expect the result to be a new value, which is why operator+ has to return a new value.1

Also note that operator+ can be slightly shortened by passing lhs by value, not by reference. However, this would be leaking implementation details, make the function signature asymmetric, and would prevent named return value optimization where result is the same object as the one being returned.

Sometimes, it's impractical to implement @ in terms of @=, such as for matrix multiplication. In that case, you can also delegate @= to @:

struct Matrix {
  // You can also define non-member functions inside the class, i.e. "hidden friends"
  friend Matrix operator*(const Matrix& lhs, const Matrix& rhs) {
    Matrix result;
    // Do matrix multiplication
    return result;
  }
  Matrix& operator*=(const Matrix& rhs)
  {
    return *this = *this * rhs; // Assuming operator= returns a reference
  }
};

The bit manipulation operators ~ & | ^ << >> should be implemented in the same way as the arithmetic operators. However, (except for overloading << and >> for output and input) there are very few reasonable use cases for overloading these.

1 Again, the lesson to be taken from this is that a += b is, in general, more efficient than a + b and should be preferred if possible.

Subscript Operator

The subscript operator is a binary operator which must be implemented as a class member. It is used for container-like types that allow access to their data elements by a key. The canonical form of providing these is this:

struct X {
        value_type& operator[](index_type idx);
  const value_type& operator[](index_type idx) const;
  // ...
};

Unless you do not want users of your class to be able to change data elements returned by operator[] (in which case you can omit the non-const variant), you should always provide both variants of the operator.

Operators for Pointer-like Types

For defining your own iterators or smart pointers, you have to overload the unary prefix dereference operator * and the binary infix pointer member access operator ->:

struct my_ptr {
        value_type& operator*();
  const value_type& operator*() const;
        value_type* operator->();
  const value_type* operator->() const;
};

Note that these, too, will almost always need both a const and a non-const version. For the -> operator, if value_type is of class (or struct or union) type, another operator->() is called recursively, until an operator->() returns a value of non-class type.

The unary address-of operator should never be overloaded.

For operator->*() see this question. It's rarely used and thus rarely ever overloaded. In fact, even iterators do not overload it.


Continue to Conversion Operators.

Kati answered 12/12, 2010 at 12:47 Comment(22)
operator->() is actually extremely weird. It's not required to return a value_type* -- in fact, it can return another class type, provided that class type has an operator->(), which will then be called subsequently. This recursive calling of operator->()s proceeds until a value_type* return type occurs. Madness! :)Rumilly
@j_random_hacker: I'm just trying to not to put too much information into these "basic rules". This answer is already very long. So I've upvoted your comment instead to make it visible above the rest of the commentary noise. However, feel free to add your information to the answer if you think it warrants being mentioned in the answer proper.Kati
Note that there are cases when defining operatorX in terms of operatorX= if not practical. For matrices and polynomials (english?), code factorization of the multiplication goes the other way: operator*= should be defined in terms of operator*.Weiler
@LucHermitte: I have yet to encounter a case where doing it the other way would be more effective, but I do not doubt that they exist. This is all rules of thumb, after all.Kati
It's not exactly about effectiveness. It's about we cannot do it in the traditional-idiomatic way in a (very) few cases: when the definition of both operands need to stay unchanged while we compute the result. And as I said, there are two classical examples: matrices multiplication, and multiplication of polynomials. We could define * in terms of *= but it would be awkward because one of the first operation of *= would to create a new object, result of the computation. Then, after the for-ijk loop, we would swap this temporary object with *this. ie. 1.copy, 2.operator*, 3.swapWeiler
Defining operatorX in terms of operatorX= prevents the former from being constexpr, making it another case of switching which one derives from the other.Vexatious
I disagree with the const/non-const versions of your pointer-like operators, e.g. ` const value_type& operator*() const;` - this would be like having a T* const returning a const T& on dereferencing, which is not the case. Or in other words: a const pointer does not imply a const pointee. In fact, it is not trivial to mimic T const * - which is the reason for the whole const_iterator stuff in the standard library. Conclusion: the signature should be reference_type operator*() const; pointer_type operator->() constIshtar
One comment: The implementation of binary arithmetic operators suggested is not such efficient as it can be. Se Boost operators headers simmetry note: boost.org/doc/libs/1_54_0/libs/utility/operators.htm#symmetry One more copy can be avoided if you use a local copy of the first parameter, do +=, and return the local copy. This enables NRVO optimization.Mama
Trying when the rule of thumb doesn't apply. The == operator treat both objects as equal, but surely it often would need to access private members to compare - will that work when it's a non-member?Thoron
@thomthom: If a class doesn't have a publicly accessible API for getting at its state, you will have to either make everything that needs to access its state a member or a friend of the class. This, of course, is also true for all operators.Kati
@jbi juanchopanzacpp.wordpress.com/2014/05/11/…Angary
I tried implementing operator| as per your suggestion to follow arithmetic rules but this doesn't compile because I have the wrong number of arguments inline X operator|(X lhs, const X &rhs) ?Seventieth
@Jon: That would be the free function implementation. You didn't put this into a class, did you? Read here: "A binary infix operator @, applied to the objects x and y, is called either as operator@(x,y) or as x.operator@(y)."Kati
@Toby Indeed. Thanks for pointing this out. However, this guide's goal is to present experience and common wisdom collected over the years, and any text about operator<=>, because of it being brand new, will not be up to that standard. But I guess that can't be helped.Kati
In your case, @JonCage, arithmetic operators such as operator|() should be defined outside the class definition, or specified as friend functions if defined inside the class definition. If defined inside the class definition, but not specified as friends, they'll be member functions (which isn't what you want).Nicety
For bitwise operators, the first use cases that come to mind are when creating a bit-field enum, or making a wrapper for a bit field.Nicety
Can someone explain "If value_type is known to refer to a built-in type, the const variant of the operator should better return a copy instead of a const reference"Mallen
It is advisable to only allow lvalues as operands of any assignment operator, which is operator= but also operator+= and friends and operator++ and operator--: Use operator+=(Rhs rhs) & and operator++() &. Doing so protects you from acting on a temporary. In C++98, you couldn't do that with members.Tramway
@MaxPower I am not active on SO anymore. I understand that this is outdated. If questions relating operator<=> are closed as a dupe of this, then that is, of course, wrong, but one of the problems that drove me away was that normal users are almost totally powerless to fix these problems. Not working on a platform providing this operator, and thus not having any experience with it, I will not add anything regarding to it. I am sorry. What would you want me to do?Kati
"Overloading unary minus and plus is not very common and probably best avoided." - why is this? Unary minus is useful for negation, especially since if one can write x - y and x + y then one should be able to write -y + x in most cases.Quarterage
Rev. 17 of this answer had a change for the binary arithmetic operators which used NRVO to keep the number of instances down to 3. That change was reverted and now it's back to 4 instances. Why was it reverted?Coacervate
isn’t it swap(*this, rhs) instead of swap(rhs)?Woolard
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550

The Three Basic Rules of Operator Overloading in C++

When it comes to operator overloading in C++, there are three basic rules you should follow. As with all such rules, there are indeed exceptions. Sometimes people have deviated from them and the outcome was not bad code, but such positive deviations are few and far between. At the very least, 99 out of 100 such deviations I have seen were unjustified. However, it might just as well have been 999 out of 1000. So you’d better stick to the following rules.

  1. Whenever the meaning of an operator is not obviously clear and undisputed, it should not be overloaded. Instead, provide a function with a well-chosen name.
    Basically, the first and foremost rule for overloading operators, at its very heart, says: Don’t do it. That might seem strange, because there is a lot to be known about operator overloading and so a lot of articles, book chapters, and other texts deal with all this. But despite this seemingly obvious evidence, there are only a surprisingly few cases where operator overloading is appropriate. The reason is that actually it is hard to understand the semantics behind the application of an operator unless the use of the operator in the application domain is well known and undisputed. Contrary to popular belief, this is hardly ever the case.

  2. Always stick to the operator’s well-known semantics.
    C++ poses no limitations on the semantics of overloaded operators. Your compiler will happily accept code that implements the binary + operator to subtract from its right operand. However, the users of such an operator would never suspect the expression a + b to subtract a from b. Of course, this supposes that the semantics of the operator in the application domain is undisputed.

  3. Always provide all out of a set of related operations.
    Operators are related to each other and to other operations. If your type supports a + b, users will expect to be able to call a += b, too. If it supports prefix increment ++a, they will expect a++ to work as well. If they can check whether a < b, they will most certainly expect to also to be able to check whether a > b. If they can copy-construct your type, they expect assignment to work as well.


Continue to The Decision between Member and Non-member.

Kati answered 12/12, 2010 at 12:45 Comment(44)
The only thing of which I am aware which violates any of these is boost::spirit lol.Goings
@Billy: According to some, abusing + for string concatenation is a violation, but it has by now become well established praxis, so that it seems natural. Although I do remember a home-brew string class I saw in the 90ies that used binary & for this purpose (referring to BASIC for established praxis). But, yeah, putting it into the std lib basically set this in stone. The same goes for abusing << and >> for IO, BTW. Why would left-shifting be the obvious output operation? Because we all learned about it when we saw our first "Hello, world!" application. And for no other reason.Kati
"the meaning of an operator is not obviously clear and undisputed" and how do you determine that?Sulphide
@curiousguy: If you have to explain it, it's not obviously clear and undisputed. Likewise if you need to discuss or defend the overloading.Kati
So overloading == for containers is an abuse?Sulphide
This seems good sense, but has drawbacks: Point 1 will make boost::spirit impossible to exist, and the same can apply to iostream (does cout << "hello world" shift bits?) Point2 literally inhibits whatever innovation: something just invented isn't "well known", but me become as such in future. std::string use + not to "add". And they where not "well known" before STL had been invented.Campanulaceous
@Emilio: These rules are for the common C++ developer. Of course they would prohibit creating anything like IO streams or spirit or, as I said above, even a decent string class. Read again, I preceded this with "Sometimes people have deviated from them and the outcome was not bad code, but such positive deviations are few and far between." IOW: When you are a beginner, strictly stick to those rules. When you are an experienced C++ developer (that's after several years), you might form your own opinion. When you deviate as far as, say, spirit does, rely on peer reviews to judge your ideas.Kati
@curiousguy: Why? What could it do but compare containers, element by element, for equality? Isn't that obvious?Kati
@sbi: "peer review" is always a good idea. To me a badly choosen operator is not different from a badly choosen function name (I saw many). Operator are just functions. No more no less. Rules are just the same. And to understand if an idea is good, the best way is understand how long does it takes to be understood. (Hence, peer review is a must, but peers must be chosen between people free from dogmas and prejudice.)Campanulaceous
@Emilio: I have no idea what this is arguing for or against. Are you sure you were replying to my comment?Kati
@Kati "Isn't that obvious?" It isn't obvious to me that two containers should compare equal even if they aren't equivalents.Sulphide
@curiousguy: That's the Java/C# approach. In C++, containers hold values, rather than references, so the difference between "equality" and "equivalence" is moot. (If I got the difference right, that is.)Kati
@Kati To me, the only absolutely obvious and indisputable fact about operator== is that it should be an equivalence relation (IOW, you should not use non signaling NaN). There are many useful equivalence relations on containers. What does equality means? "a equals b" means that a and b have the same mathematical value. The concept of mathematical value of a (non-NaN) float is clear, but the mathematical value of a container can have many distinct (type recursive) useful definitions. The strongest definition of equality is "they are the same objects", and it is useless.Sulphide
(...) The second strongest definition is "holds the same objects", and it is useless for value containers. Another strong definition is "two objects are equal iff they cannot be distinguished using their public interface" (comparison of addresses isn't "part of the public interface"). This relates to simulation and bi-simulation. The nuances with this definition comes with the parts of the interface you are allowed to use.Sulphide
(...) Are the two set<int,function<bool(int,int)>> ({1},<) and ({1},>) equal? They have the same elements, but 1) if you can "compare" their comparator, you will be able to distinguish them; 2) if you add another element 2, their elements as a sequence won't be the same. Are two hash_set<int> with the same elements in a different order equal? Neither of these questions are absolutely obvious (at least to me).Sulphide
(...) Also, what should operator= (x) do? Should it be equivalent with copy construction? Should it be equivalent with assign(x.begin(),x.end())? Both are very reasonable choices. I feel that there are strong arguments for both. This seems to argue against providing this operator at all.Sulphide
@sbi: It's conceptually against, but recognizes a good point. Essentially I don't believe in a difference between operators and functions: if + for "concatenation" or unary* for "keen star" or ~for "orthogonal" or "transposed" can be questionable, std::string::empty is also questionable (an imperative verb used to check a state ?!? Wouldn't it be is_empty better?). (...)Campanulaceous
(...) In any case, wherever you define a symbolic name, you define a convention between that name and what it represent. This convention cannot be obvious to anyone but you, until you don't expose enough "information" (in "information theory" sense) to your users (may be a comment, a document or a trivial use case, or the raw source code- when very simple). But I must see a clean contradiction (and may be even a "clutural racism" -note the quotes- in the background: saying "When you are a beginner, strictly stick to those rules", will defeat any geniality to emerge. (...)Campanulaceous
(...) The real fact is that, when you are a beginner the first sample of overloading yopu see is cout << "hello world", that is just a violation to the rules (<< should shift bits!) that is hidden by teacher lies presenting it a the "insertion operator". Now 1 only can be true: 1) the "insertion operator" is a lye, and is a violation to the rules thaught to the very first lesson about rules (LOL!) or ... 2) << is just a symbol whose meaning depends top what it is applied to, and hence no prejudice should exist about its meaning. (...)Campanulaceous
(...) The "insertion operator" is for stream and the shift operator for integers. The both have incidentally the same aspect. For this reason I always teach about << as "double left arrow" and + about "crosss". Whether "cross" becomes "add" or "concat", or "merge" etc., is something I have to teach when talking about the type it is related. Sorry to anyone for the very long comment.Campanulaceous
@curiousguy: No, you are making this way to complicated. The obvious implementation for operator==() on a heterogeneous container (aka struct) is that, for all of its elements (aka "members"), operator==() returns true when applied to the corresponding element in another struct of the same type. The obvious meaning for operator==() of a non-heterogeneous container is that, for all elements in the container, operator==() returns true when applied to the corresponding object in another container of the same type.Kati
@Emilio: I have allowed for violations of the rules, and I have listed IO streams and strings (and spirit) as obvious examples of those violations being successful. But that doesn't mean that a C++ newbie thinking overloading ! for strings might be a neat idea has it right. IO streams were designed ~30 years ago, when operator overloading was the newest thing in C++, brings a lot of historical weight, and, as you pointed out, newbies will usually learn the meaning of << as output operator before its meaning as a shift operator.Kati
@Emilio: std::string, despite all its shortcomings, unified the C++ world which was split into thousands of different string classes. That alone was a great deed, and the designers would have been hard-pressed to do anything else so wrong for it to not be seen as a success — despite the fact that some disagreed with its (ab)use of + for concatenation.Kati
@Emilio: And spirit deliberately set out to abuse operator overloading in order to create an embedded domain-specific language. That's not necessarily the first thing a newbie does, it might well be one of the most often legitimate reasons to abuse a programming language, and a very hard task to succeed with unless you have the backup of a strong community.Kati
@sbi: Your all arguments reduce to "don't do unless somebody else already did it a number of times". Apply this to "make children" and to "anyone", and humanity will extinguish in one generation time! It's a hard dirty work, but someone has to begin. Newbie or experts, is not the point. The idea is the point. If there is enough "believers" the idea goes, otherwise will die. Don't be "racist" against newbies: instead to tell them to "go away from water", teach them "how to swim"! (You may be surprised!)Campanulaceous
@Emilio: Yes, my suggestion to those seeking advice is to not to invent something new. In the last 20 years, I have seen a lot of abuse of operator overloading, and very little justified use. (And I have been guilty of this, too, in my first few years of C++.) Consequently, my advice is to stick to well-known paradigms, unless you know enough of operator overloading that you wouldn't need advice in the first place. (As for the rest of your argument: The comparisons fall so awkwardly short, I won't spend time on them.)Kati
Please take extended discussions to Stack Overflow Chat, in particular this roomErrick
@Kati So, are you trying to say that operator== for containers is ill-specified, or are you just saying it without actually trying to say that?Sulphide
@Kati "In the last 20 years, I have seen a lot of abuse of operator overloading, and very little justified use." What kind of abuses have you seen? Mostly subtle abuses, or ridiculous abuses?Sulphide
@BenVoigt after x = y; the value/state of x is equal to/equivalent with the value/state of y before, so it is not totally crazy. Also, the C++ language allows both C(const C&) and C(C&) as copy-ctor signatures, and operator=(const C&) as well as operator=(C&) for assignment operator signature, so the core language quite explicitly supported this design too.Sulphide
@curiousguy: The language allows defining operator+ as subtraction also. It's a very VERY bad idea, because it's surprising. x = y; should not change y. It does not do so for any built-in type. Support for C::operator=(C&) is needed to support things like reference-counted pointers, where internal state may change but the user-visible state does. std::auto_ptr is universally recognized as a mistake.Winnebago
@BenVoigt "The language allows defining operator+ as subtraction also." actually the code language doesn't know addition from subtraction, so it couldn't tell. OTOH, the core language knows quite well what a copy ctor means. "Support for C::operator=(C&) is needed to support things like reference-counted pointers, where internal state may change but the user-visible state does." No it is not.Sulphide
Yeah, I agree that cout << "Hello world" is the worst violation of these rules. I'm convinced that it was implemented as a demonstration of what operator overloading could do, and we never intended to become used in the real world.Plumb
Why did you write that Comparison operators should be implemented as non-member functions? What is the reason?Jaimiejain
@Sekory: Did I even write this in this answer? Anyway, it's explained here: If a binary operator treats both operands equally (it leaves them unchanged), implement this operator as a non-member function. What exactly do you want to know.Kati
With respect to "whenever the meaning of an operator is not obviously clear and undisputed, it should not be overloaded", would overloading < and > for a vector class comparing length violate this principle?Lindstrom
@Robinson: The STL defines comparison of containers as lexicographical comparison, not as length. Therefore, everybody looking at v1 < v2 would expect it to do so, too. For what you want, there's v1.size() < v2.size().Kati
The assignment operator is overloaded by default, right?Thoroughfare
@Moiz: In case you want to know whether the compiler generates an assignment operator by default, see this question and its answer.Kati
Tying into these rules, I would personally say that the best reason to overload an operator is to provide consistency with similar classes. If you're making a math-related class, for instance, it would be reasonable to overload assignment, arithmetic, and comparison operators for use with said class, conversion operators to convert to built-in and/or standard math types, and I/O operators; not so much array, new/delete, or pointer-like operators. Conversely, if you're making a video game character class, it's unreasonable to overload arithmetic operators to modify the character's level.Nicety
How would you feel about operator| for piping, such as in range adaptors?Dexterdexterity
@Daniel: Look, if nobody had come up with overloading << and >> for IO, and you came up with it now, it would be frowned upon. Since it's been done 30 years ago, it's part of the domain now. The same goes for everything else that's not in common use: If you or I do it in our projects, it's rather likely our co-workers will complain. If it is done in a popular boost library, and fits the domain of the code (using | for piping is well-known), this might be different.Kati
The real world is not that ideal. For example, the operator+ should be Abelian (i.e. commutative) to serve to the well-formed semantics, but almost every design of string concatenation by operator overloading break this rule.Neva
@Neva See this comment upthread: stackoverflow.com/questions/4421706/…Kati
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292

The Decision between Member and Non-member

Category Operators Decision
Mandatory member functions [], (), =, ->, ... member function (mandated by the C++ standard)
Pointer-to-member access ->* member function
Unary ++, -, *, new, ... member function, except for enumerations
Compound assignment +=, |=, *=, ... member function, except for enumerations
Other operators +, ==, <=>, /, ... prefer non-member

The binary operators = (assignment), [] (array subscription), -> (member access), as well as the n-ary () (function call) operator, must always be implemented as member functions, because the syntax of the language requires them to.

Other operators can be implemented either as members or as non-members. Some of them, however, usually have to be implemented as non-member functions, because their left operand cannot be modified by you. The most prominent of these are the input and output operators << and >>, whose left operands are stream classes from the standard library which you cannot change.

For all operators where you have to choose to either implement them as a member function or a non-member function, use the following rules of thumb to decide:

  1. If it is a unary operator, implement it as a member function.
  2. If a binary operator treats both operands equally (it leaves them unchanged), implement this operator as a non-member function.
  3. If a binary operator does not treat both of its operands equally (usually it will change its left operand), it might be useful to make it a member function of its left operand’s type, if it has to access the operand's private parts.

Of course, as with all rules of thumb, there are exceptions. If you have a type

enum Month {Jan, Feb, ..., Nov, Dec}

and you want to overload the increment and decrement operators for it, you cannot do this as a member functions, since in C++, enum types cannot have member functions. So you have to overload it as a free function. And operator<() for a class template nested within a class template is much easier to write and read when done as a member function inline in the class definition. But these are indeed rare exceptions.

(However, if you make an exception, do not forget the issue of const-ness for the operand that, for member functions, becomes the implicit this argument. If the operator as a non-member function would take its left-most argument as a const reference, the same operator as a member function needs to have a const at the end to make *this a const reference.)


Continue to Common operators to overload.

Kati answered 12/12, 2010 at 12:49 Comment(34)
Herb Sutter's item in Effective C++ (or is it C++ Coding Standards?) says one should prefer non-member non-friend functions to member functions, to increase the encapsulation of the class. IMHO, the encapsulation reason takes precedence to your rule of thumb, but it does not decrease the quality value of your rule of thumb.Frottage
@paercebal: Effective C++ is by Meyers, C++ Coding Standards by Sutter. Which one are you referring to? Anyway, I dislike the idea of, say, operator+=() not being a member. It has to change its left-hand operand, so by definition it has to dig deep into its innards. What would you gain by not making it a member?Kati
@Frottage You are referring to Effective C++ by Meyers, not Sutter.Foursquare
@Polybos: Which item is that in EC++?Kati
@sbi: Item 44 in C++ Coding Standards (Sutter) Prefer writing nonmember nonfriend functions, of course, it only applies if you can actually write this function using only the public interface of the class. If you cannot (or can but it would hinder performance badly), then you have to make it either member or friend.Aristophanes
@Kati : Oops, Effective, Exceptional... No wonder I mix the names up. Anyway the gain is to limit as much as possible the number of functions that have access to an object private/protected data. This way, you increase the encapsulation of your class, making its maintenance/testing/evolution easier.Frottage
@Kati : One example. Let's say you're coding a String class, with both the operator += and the append methods. The append method is more complete, because you can append a substring of the parameter from index i to index n -1: append(string, start, end) It seems logical to have += call append with start = 0 and end = string.size. At that moment, append could be a member method, but operator += doesn't need to be a member, and making it a non-member would decrease the quantity of code playing with the String innards, so it is a good thing.... ^_^ ...Frottage
@paercebal: (Did you know there's also Efficient C++?) I see. If you had an append() method to your string class, += could indeed be a non-member. But, for one, using + and += for strings is questionable at best (isn't + supposed to b be commutative?) and only alright because it's based on existing praxis. Also, even though I do agree with Sutter on this (BTW, I think it was Meyers who first published an article on non-members actually improving encapsulation, hence my confusion) I might still make += a member for the very same reason: existing praxis.Kati
@Kati : I did know about both Sutter and Meyers books: I own almost all of them.Frottage
@Kati : Commutativity should not be a criterion. For example, Matrices multiplication is not commutative, while number multiplication is. Should I be forbidden to overload * for matrices just because the C++ native integer * is commutative? No. Here, the principle of least surprise apply, and we must analyze an operator according to its context (i.e. its parameters). The same goes for strings and operator + (and by extension, +=). The fact is, no one expects the + operator to be commutative on strings, so, where's the problem with that?Frottage
@Kati : Now, the existing practice about an operator could apply to an interface (e.g. provide + and += operators on strings because users do expect them), it should not be used to decide the implementation detail IMHO (e.g. decide += is to be member or non-member). This implementation detail is not about personal preference of taste: Decreasing the quantity of code accessing private data (i.e. encapsulation) is a good thing, and can be measured, so it should be pursued as much as possible (as long as the code remains clear, readable and correct, of course)Frottage
It's definitely better to implement either + or += outside the class (and likewise for the other operators @ that come with a @= version) for the reason paercebal gave. Here's an idea: since += is usually more efficient, implement += inside the class, and then have a templated operator+() that is enabled using enable_if only for classes T for which want_helper_ops<T> is defined. Nuke that boilerplate! :)Rumilly
@paercebal: Yes, using << and >> for IO and + for string concatenation is established praxis. And I'm not opposed to any of them. What I'm saying is that there are arguments against doing so, and that, until it is established praxis, according my Three Basic Rules, to it shouldn't be done. (Yes I know it can't become established praxis unless you violate rule #1.) As for making += a non-member: Yes, it seems logical, but it never occurred to me, because I overload operators by those three rules for more than a decade. And such long-nursed habits die hard.Kati
@j_random_hacker: The question is not whether one of them should be non-member, I think that's unanimously clear. The question is whether both of them should. See me previous comment to paercebal regarding this.Kati
This rule sounds weird: If a binary operator does not treat both of its operands equally (usually it will change its left operand), it should be a member function of its left operand’s type.. The motivation for making it a member is not that both operands aren't treated equal. I think you should reword that item. An operator/ that does numeric division will treat the left operand different from the right. Still it should be written as a non-member.Bitolj
@Kati : "And such long-nursed habits die hard." : I know... I'm trying to get rid of my hungarian notation habit... :-(Frottage
@Johannes: If an operator changes its left-hand operand, it needs access to its innards.Kati
@Kati : "If an operator changes its left-hand operand, it needs access to its innards". No: The operator could use a method, so it doesn't need to have access to protected/private member. Again, in my example for strings, operator += doesn't need to access private members. It only needs only call the public append method with the right parameters.Frottage
@paercebal: I changed the wording somewhat. Do you approve of this now or does it need further changes?Kati
@Kati : After thinking about it, the list should be something like: 1. "if possible, make it non-member non-friend", 2. "The following operators are available only as member-functions: ...". . . Everything else seems a matter of taste. My own viewpoint on the subject is similar to yours, but it is still a matter of taste, and this excellent question/answers set should not be tainted by personal tastes. It should be the reference, motivated only by rock hard reasons.Frottage
@paercebal: But I consider "if possible..." too vague for a rule of thumb. And that's what I'm trying to hand out here: an easy-to-remember rule of thumb, which you can remember even if you can't remember all the reasoning. As it now is, my rule is basically saying "make it a non-member, if it's changing the left operand, consider membership". If most of you out there don't think this is enough, I will change t. But "if possible..." isn't enough to improve what I now have.Kati
@Kati : For me, the "if possible..." means "if it compiles...". Some operators must be member functions. For the others, I guess non-member functions should be preferred anyway, even if friend, because these authorizes a cast from some value into the class value (e.g. having the code MyInteger m(42), n ; n = 25 + m ; work).Frottage
@Kati : With your authorization, I could think about an alternate list of items, and write it as a "proposition" at the end of the post (the comments are not suited for this lengthy discussion). Another possibility would be to contact me by email (append "@" + "gmail" + "." + "com" to my identifier)Frottage
@paercebal: If it's too long for comments, why don't we discuss this in the C++ chat? That's where the whole FAQ idea started out. It's pretty quiet there now (and I can't do online discussions now either), but in about 5-10hrs it usually looks very different. (If you want, you can post your ideas there now anyway, they won't get lost.)Kati
@Kati I only just saw this thread and I hope this question will be answered though this thread is nearly 2 years old now. With respect to overloading the binary operators, why should the operator not be a member function if it treats both operands equally (or doesn't change their innards)?Fishhook
@AndrewFalanga this comment answers your question: https://mcmap.net/q/15363/-what-are-the-basic-rules-and-idioms-for-operator-overloading/…Biscay
Is there a recommended place for helper (non-member) functions? In the same file as the class, or in a separate file but in the same namespace as the class?Biscay
@Dennis: If they are part of the class' interface (like widget operator+(const widget&, const widget&) or void normalize(widget&), they should be in the class' namespace and close to the class' declaration. If they are merely private helpers, hide them in (the unnamed namespace of) the .cpp file or, should everything be in headers, in some namespace details.Kati
I may be wrong, but I feel that the case of the unary minus operator is forgotten (-a)… The rules of thumb states that “If it is a unary operator, implement it as a member function.”, but unary minus operator is a unary operator, yet it would be more logical to implement this operator as a non-member function (because after all, it leaves the operand unchanged). By the way, the same holds for unary plus operator.Sadonia
@mlpo: As far as I can see I hadn't even mentioned unary minus anywhere except on the list of operators to overload. That's clearly an oversight, caused by unary minus not being very useful to overload at all. In fact, I don't believe that among the horrible abuse of operator overloading I have seen in >20 years was a single case of an overloaded unary minus operator. Nevertheless, thanks for pointing this out. I have now at least mentioned them. As per the rules, I'd made them members. Seems fitting to me.Kati
@sbi, the unary minus operator can sometimes be useful when adding two objects is really complex, we write the addition once and then we can implement the subtraction operator by writing a + (-b). This prevents code replication.Sadonia
Furthermore, I am not sure that the rule is logical, because unary minus operator leaves the operand unchanged and therefore in the same way as addition, it should be non-member function, right?Sadonia
@mlpo: 1. You could just as well implement the subtraction with a.data + (-b.data) without adding a minus operator to the class. (Of course, if the class is supposed to be a number type, then operator-() (and operator+()) should be part of the interface as per rule #3 here.) 2. As per rule #1 above unary operators should be members, no matter whether they change the object. (Besides prefix increment/decrement I can't think of any that do.) That's because you rarely want to conversions etc. applied to them.Kati
@Frottage There are some missing points. 1. friend contributes to ADL-only lookup, hence less namespace-scope pollution. This works in despite of the existence of non-public members. So, making something non-member as possible is just not that idiomatic. 2. A simple overload of non-member operator+= fails to work on prvalues allowing modifications, but a member overload just works. 3. Diagnostics on non-member overloading failures can be a disaster. Usually not that bad for members.Neva
K
291

The General Syntax of operator overloading in C++

You cannot change the meaning of operators for built-in types in C++, operators can only be overloaded for user-defined types1. That is, at least one of the operands has to be of a user-defined type. As with other overloaded functions, operators can be overloaded for a certain set of parameters only once.

Not all operators can be overloaded in C++. Among the operators that cannot be overloaded are: . :: sizeof typeid .* and the only ternary operator in C++, ?:

Among the operators that can be overloaded in C++ are these:

Category Operators Arity and Placement
Arithmetic + - * / % and += -= *= /= %= binary infix
+ - unary prefix
++ -- unary prefix and postfix
Bitwise & | ^ << >> and &= |= ^= <<= >>= binary infix
~ unary prefix
Comparison == != < > <= >= <=> binary infix
Logical || && binary infix
! unary prefix
Allocation functions new new[] delete delete[] unary prefix
User-defined conversions T unary
Assignment = binary infix
Member access -> ->* binary infix
Indirection/Address-of * & unary prefix
Function call () N-ary postfix
Subscript [] N-ary2 postfix
Coroutine await co_await unary prefix
Comma , binary infix

However, the fact that you can overload all of these does not mean you should do so. See the next answer.

In C++, operators are overloaded in the form of functions with special names. As with other functions, overloaded operators can generally be implemented either as a member function of their left operand's type or as non-member functions. Whether you are free to choose or bound to use either one depends on several criteria.3 A unary operator @4, applied to an object x, is invoked either as operator@(x) or as x.operator@(). A binary infix operator @, applied to the objects x and y, is called either as operator@(x,y) or as x.operator@(y).5

Operators that are implemented as non-member functions are sometimes friend of their operand’s type.

1 The term “user-defined” might be slightly misleading. C++ makes the distinction between built-in types and user-defined types. To the former belong for example int, char, and double; to the latter belong all struct, class, union, and enum types, including those from the standard library, even though they are not, as such, defined by users.

2 The subscript operator used to be binary, not N-ary until C++23.

3 This is covered in a later part of this FAQ.

4 The @ is not a valid operator in C++ which is why I use it as a placeholder.

5 The only ternary operator in C++ cannot be overloaded and the only n-ary operator must always be implemented as a member function.


Continue to The Three Basic Rules of Operator Overloading in C++.

Kati answered 12/12, 2010 at 12:46 Comment(1)
Addendum: Some compilers provide a (typically hidden, internal) switch that allows you to overload operators for non-user-defined types. Any code that redefines built-in operators in this manner is non-compliant.Nicety
G
189

Conversion Operators (also known as User Defined Conversions)

In C++ you can create conversion operators, operators that allow the compiler to convert between your types and other defined types. There are two types of conversion operators, implicit and explicit ones.

Implicit Conversion Operators (C++98/C++03 and C++11)

An implicit conversion operator allows the compiler to implicitly convert (like the conversion between int and long) the value of a user-defined type to some other type.

The following is a simple class with an implicit conversion operator:

class my_string {
public:
  operator const char*() const {return data_;} // This is the conversion operator
private:
  const char* data_;
};

Implicit conversion operators, like one-argument constructors, are user-defined conversions. Compilers will grant one user-defined conversion when trying to match a call to an overloaded function.

void f(const char*);

my_string str;
f(str); // same as f( str.operator const char*() )

At first this seems very helpful, but the problem with this is that the implicit conversion even kicks in when it isn’t expected to. In the following code, void f(const char*) will be called because my_string() is not an lvalue, so the first does not match:

void f(my_string&);
void f(const char*);

f(my_string());

Beginners easily get this wrong and even experienced C++ programmers are sometimes surprised because the compiler picks an overload they didn’t suspect. These problems can be mitigated by explicit conversion operators.

Explicit Conversion Operators (C++11)

Unlike implicit conversion operators, explicit conversion operators will never kick in when you don't expect them to. The following is a simple class with an explicit conversion operator:

class my_string {
public:
  explicit operator const char*() const {return data_;}
private:
  const char* data_;
};

Notice the explicit. Now when you try to execute the unexpected code from the implicit conversion operators, you get a compiler error:

prog.cpp: In function ‘int main()’:
prog.cpp:15:18: error: no matching function for call to ‘f(my_string)’
prog.cpp:15:18: note: candidates are:
prog.cpp:11:10: note: void f(my_string&)
prog.cpp:11:10: note:   no known conversion for argument 1 from ‘my_string’ to ‘my_string&’
prog.cpp:12:10: note: void f(const char*)
prog.cpp:12:10: note:   no known conversion for argument 1 from ‘my_string’ to ‘const char*’

To invoke the explicit cast operator, you have to use static_cast, a C-style cast, or a constructor style cast ( i.e. T(value) ).

However, there is one exception to this: The compiler is allowed to implicitly convert to bool. In addition, the compiler is not allowed to do another implicit conversion after it converts to bool (a compiler is allowed to do 2 implicit conversions at a time, but only 1 user-defined conversion at max).

Because the compiler will not cast "past" bool, explicit conversion operators now remove the need for the Safe Bool idiom. For example, smart pointers before C++11 used the Safe Bool idiom to prevent conversions to integral types. In C++11, the smart pointers use an explicit operator instead because the compiler is not allowed to implicitly convert to an integral type after it explicitly converted a type to bool.

Continue to Overloading new and delete.

Gaskins answered 17/5, 2013 at 18:32 Comment(0)
K
167

Overloading new and delete operators

Note: This only deals with the syntax of overloading new and delete, not with the implementation of such overloaded operators. I think that the semantics of overloading new and delete deserve their own FAQ, within the topic of operator overloading I can never do it justice.

Basics

In C++, when you write a new expression like new T(arg) two things happen when this expression is evaluated: First operator new is invoked to obtain raw memory, and then the appropriate constructor of T is invoked to turn this raw memory into a valid object. Likewise, when you delete an object, first its destructor is called, and then the memory is returned to operator delete.
C++ allows you to tune both of these operations: memory management and the construction/destruction of the object at the allocated memory. The latter is done by writing constructors and destructors for a class. Fine-tuning memory management is done by writing your own operator new and operator delete.

The first of the basic rules of operator overloading – don’t do it – applies especially to overloading new and delete. Almost the only reasons to overload these operators are performance problems and memory constraints, and in many cases, other actions, like changes to the algorithms used, will provide a much higher cost/gain ratio than attempting to tweak memory management.

The C++ standard library comes with a set of predefined new and delete operators. The most important ones are these:

void* operator new(std::size_t) throw(std::bad_alloc); 
void  operator delete(void*) throw(); 
void* operator new[](std::size_t) throw(std::bad_alloc); 
void  operator delete[](void*) throw(); 

The first two allocate/deallocate memory for an object, the latter two for an array of objects. If you provide your own versions of these, they will not overload, but replace the ones from the standard library.
If you overload operator new, you should always also overload the matching operator delete, even if you never intend to call it. The reason is that, if a constructor throws during the evaluation of a new expression, the run-time system will return the memory to the operator delete matching the operator new that was called to allocate the memory to create the object in. If you do not provide a matching operator delete, the default one is called, which is almost always wrong.
If you overload new and delete, you should consider overloading the array variants, too.

Placement new

C++ allows new and delete operators to take additional arguments.
So-called placement new allows you to create an object at a certain address which is passed to:

class X { /* ... */ };
char buffer[ sizeof(X) ];
void f()
{ 
  X* p = new(buffer) X(/*...*/);
  // ... 
  p->~X(); // call destructor 
} 

The standard library comes with the appropriate overloads of the new and delete operators for this:

void* operator new(std::size_t,void* p) throw(std::bad_alloc); 
void  operator delete(void* p,void*) throw(); 
void* operator new[](std::size_t,void* p) throw(std::bad_alloc); 
void  operator delete[](void* p,void*) throw(); 

Note that, in the example code for placement new given above, operator delete is never called, unless the constructor of X throws an exception.

You can also overload new and delete with other arguments. As with the additional argument for placement new, these arguments are also listed within parentheses after the keyword new. Merely for historical reasons, such variants are often also called placement new, even if their arguments are not for placing an object at a specific address.

Class-specific new and delete

Most commonly you will want to fine-tune memory management because measurement has shown that instances of a specific class, or of a group of related classes, are created and destroyed often and that the default memory management of the run-time system, tuned for general performance, deals inefficiently in this specific case. To improve this, you can overload new and delete for a specific class:

class my_class { 
  public: 
    // ... 
    void* operator new(std::size_t);
    void  operator delete(void*);
    void* operator new[](std::size_t);
    void  operator delete[](void*);
    // ...  
}; 

Overloaded thus, new and delete behave like static member functions. For objects of my_class, the std::size_t argument will always be sizeof(my_class). However, these operators are also called for dynamically allocated objects of derived classes, in which case it might be greater than that.

Global new and delete

To overload the global new and delete, simply replace the pre-defined operators of the standard library with our own. However, this rarely ever needs to be done.

Kati answered 12/12, 2010 at 13:7 Comment(5)
What about nothrow new ? The rule of thumb is that whenever you overload new, you must write 12 functions: [array] [{ placement | nothrow }] { new | delete }.Hamlin
@Alexandre: This whole FAQ entry is distilled from guidelines I use for teaching C++, and since nothrow new is more or less just for legacy stuff, I haven't ever even mentioned it. What would you suggest should be written about it here?Kati
@sbi: See (or better, link to) gotw.ca/publications/mill15.htm . It is only good practice towards people which sometimes use nothrow new.Hamlin
new(buffer) X(/*...*/) what actually guarantees that buffer is properly aligned for X?Sulphide
"If you do not provide a matching operator delete, the default one is called" -> Actually, if you add any arguments and do not create a matching delete, no operator delete is called at all, and you have a memory leak. (15.2.2, the storage occupied by the object is deallocated only if an appropriate ... operator delete is found)Pierre
U
57

Why can't operator<< function for streaming objects to std::cout or to a file be a member function?

Let's say you have:

struct Foo
{
   int a;
   double b;

   std::ostream& operator<<(std::ostream& out) const
   {
      return out << a << " " << b;
   }
};

Given that, you cannot use:

Foo f = {10, 20.0};
std::cout << f;

Since operator<< is overloaded as a member function of Foo, the LHS of the operator must be a Foo object. Which means, you will be required to use:

Foo f = {10, 20.0};
f << std::cout

which is very non-intuitive.

If you define it as a non-member function,

struct Foo
{
   int a;
   double b;
};

std::ostream& operator<<(std::ostream& out, Foo const& f)
{
   return out << f.a << " " << f.b;
}

You will be able to use:

Foo f = {10, 20.0};
std::cout << f;

which is very intuitive.

Uniliteral answered 22/1, 2016 at 19:0 Comment(1)
I ve always read this regarding overloading << to be used with std::cout but what about overloading << to be used with the same class. In that case it can be member function right?Alisealisen
S
6

Comparison Operators, including Three-Way Comparison(C++20)

There are equality comparisons == and !=, and relational comparisons <, >, <=, >=. C++20 has also introduced the three-way comparison operator <=>.

Operator Meaning and Notes (Old) Meaning and Notes (C++20)
x == y true if x and y are equal

satisfies EqualityComparable
(used by std::unordered_map)
(x <=> y) == 0
(usually implemented directly, not
delegated to three-way unless = default)

satisfies std::equality_comparable
x != y !(x == y) !(x == y)
x < y true if x is lower than y

satisfies LessThanComparable
(used by std::set, std::sort, etc.
but requires strict weak ordering)
(x <=> y) < 0

may satisfy std::strict_weak_ordering
when wrapped in a functor
(e.g. std::ranges::less)
x > y y < x (x <=> y) > 0
x <= y !(x < y) for strong orderings,
x == y || x < y otherwise
(x <=> y) <= 0
x >= y y <= x (x <=> y) >= 0
x <=> y N/A three-way comparison
aka. "spaceship operator"

satisfies std::three_way_comparable

Guidelines

  1. Comparison operators shouldn't be member functions.1)
  2. If you define ==, define != too (unless it is rewritten in C++20).
  3. If you define <, define >, <=, and >= too.
  4. (C++20) Prefer defining <=> over defining each relational operator.
  5. (C++20) Prefer defaulting operators over implementing manually.
  6. Equality and relational comparisons should match, meaning that
    x == y should be equivalent to !(x < y) && !(y < x)2)
  7. Don't define == in terms of <, even when you could 3)

1) Otherwise, implicit conversions would be asymmetrical, and == is expected to apply the same kinds of implicit conversions to both sides.
2) This equivalence does not apply to float, but does apply to int and other strongly ordered types.
3) This is motivated by readability, correctness, and performance.

Implementation and Common Idioms Prior to C++20

Disclaimer
If you're using C++20, the implementations in this section have been obsoleted.
Skip ahead to the C++20 parts unless you're interested in a historical perspective.

All operators are typically implemented as non-member functions, possibly as hidden friends (friends where the function is defined inside the class). All following code examples use hidden friends because this becomes necessary if you need to compare private members anyway.

struct S {
    int x, y, z;

    // (In)equality comparison:
    // implementing a member-wise equality
    friend bool operator==(const S& l, const S& r) {
        return l.x == r.x && l.y == r.y && l.z == r.z;
    }
    friend bool operator!=(const S& l, const S& r) { return !(l == r); }

    // Relational comparisons:
    // implementing a lexicographical comparison which induces a
    // strict weak ordering.
    friend bool operator<(const S& l, const S& r) {
        if (l.x < r.x) return true;   // notice how all sub-comparisons
        if (r.x < l.x) return false;  // are implemented in terms of <
        if (l.y < r.y) return true;
        if (r.y < l.y) return false; // also see below for a possibly simpler
        return l.z < r.z;            // implementation
    }
    friend bool operator>(const S& l, const S& r) { return r < l; }
    friend bool operator<=(const S& l, const S& r) { return !(r < l); }
    friend bool operator>=(const S& l, const S& r) { return !(l < r); }
};

Note: in C++11, all of these can typically be noexcept and constexpr.

Implementing all relational comparisons in terms of < is not valid if we have a partially ordered member (e.g. float). In that case, <= and >= must be written differently.

friend bool operator<=(const S& l, const S& r) { return l == r || l < r; }
friend bool operator>=(const S& l, const S& r) { return r <= l; }

Further Notes on operator<

The implementation of operator< is not so simple because a proper lexicographical comparison cannot simply compare each member once. {1, 2} < {3, 0} should be true, even though 2 < 0 is false.

A lexicographical comparison is a simple way of implementing a strict weak ordering, which is needed for containers like std::set and algorithms like std::sort. In short, a strict weak ordering should behave like the < operator for integers, except that some integers are allowed to be equivalent (e.g. for all even integers, x < y is false).

If x != y is equivalent to x < y || y < x, a simpler approach is possible:

friend bool operator<(const S& l, const S& r) {
    if (l.x != r.x) return l.x < r.x;
    if (l.y != r.y) return l.y < r.y;
    return l.z < r.z;
}

Common Idioms

For multiple members, you can use std::tie to implement comparison lexicographically:

#include <tuple>

struct S {
    int x, y, z;

    friend bool operator<(const S& l, const S& r) {
        return std::tie(l.x, l.y, l.z) < std::tie(r.x, r.y, r.z);
    }
};

Use std::lexicographical_compare for array members.

Some people use macros or the curiously recurring template pattern (CRTP) to save the boilerplate of delegating !=, >, >=, and <=, or to imitate C++20's three-way comparison.

It is also possible to use std::rel_ops (deprecated in C++20) to delegate !=, >, <=, and >= to < and == for all types in some scope.


Default Comparisons (C++20)

A substantial amount of comparison operators simply compare each member of a class. If so, the implementation is pure boilerplate and we can let the compiler do it all:

struct S {
    int x, y, z;
    // ==, !=, <, >, <=, >= are all defined.
    // constexpr and noexcept are inferred automatically.
    friend auto operator<=>(const S&, const S&) = default;
};

Note: defaulted comparison operators need to be friends of the class, and the easiest way to accomplish that is by defining them as defaulted inside the class. This makes them "hidden friends".

Alternatively, we can default individual comparison operators. This is useful if we want to define equality comparison, or only relational comparison:

friend bool operator==(const S&, const S&) = default; // inside S

See the cppreference article on default comparison.

Expression Rewriting (C++20)

In C++20, if a comparison isn't directly implemented, the compiler will also try to use rewrite candidates. Thanks to this, even if <=> isn't defaulted (which would implement all operators), we only have to implement == and <=>, and all other comparisons are rewritten in terms of these two.

Operator Potential Rewrites
x == y y == x
x != y !(x == y) or !(y == x) if equality comparison returns bool
x < y (x <=> y) < 0 or 0 < (y <=> x) if comparison result is comparable to zero
x > y (x <=> y) > 0 or 0 > (y <=> x) if ...
x <= y (x <=> y) <= 0 or 0 <= (y <=> x) if ...
x >= y (x <=> y) >= 0 or 0 >= (y <=> x) if ...
struct S {
    int x, y, z;
    // ==, !=
    friend constexpr bool operator==(const S& l, const S& r) noexcept { /* ... */ }
    // <=>, <, >, <=, >=
    friend constexpr auto operator<=>(const S& l, const S& r) noexcept { /* ... */ }
};

Note: constexpr and noexcept are optional, but can almost always be applied to comparison operators.

Three-Way Comparison Operator (C++20)

Note: it is colloquially called "spaceship operator". See also .

The basic idea behind x <=> y is that the result tells us whether x is lower than, greater than, equivalent to, or unordered with y. This is similar to functions like strcmp in C.

// old C style
int compare(int x, int y) {
    if (x < y) return -1;
    if (x > y) return  1;
    return             0; // or simply return (x > y) - (x < y);
}
// C++20 style: this is what <=> does for int.
auto compare_cxx20(int x, int y) {
    if (x < y) return std::strong_ordering::less;
    if (x > y) return std::strong_ordering::greater;
    return            std::strong_ordering::equal;
}
// This is what <=> does for float.
auto compare_cxx20(float x, float y) {
    if (x < y)  return std::partial_ordering::less;
    if (x > y)  return std::partial_ordering::greater;
    if (x == y) return std::partial_ordering::equivalent;
    return             std::partial_ordering::unordered; // NaN
}

Comparison Categories

The result of this operator is neither bool nor int, but a value of comparison category.

Comparison Category Example Possible Values
std::strong_ordering int less, equal = equivalent, greater
std::weak_ordering user-defined1) less, equivalent, greater
std::partial_ordering float less, equivalent, greater, unordered

std::strong_orderings can be converted to std::weak_ordering, which can be converted to std::partial_ordering. Values of these categories are comparable to (e.g. (x <=> y) == 0) and this has similar meaning to the compare function above. However, std::partial_ordering::unordered returns false for all comparisons.


1) There are no fundamental types for which x <=> y results in std::weak_ordering. Strong and weak orderings are interchangeable in practice; see Practical meaning of std::strong_ordering and std::weak_ordering.

Manual Implementation of Three-Way Comparison

Three-way comparison is often defaulted, but could be implemented manually like:

#include <compare> // necessary, even if we don't use std::is_eq

struct S {
    int x, y, z;
    // This implementation is the same as what the compiler would do
    // if we defaulted <=> with = default;
    friend constexpr auto operator<=>(const S& l, const S& r) noexcept {
        // C++17 if statement with declaration makes this more readable.
        // !std::is_eq(c) is not the same as std::is_neq(c); it is also true
        // for std::partial_order::unordered.
        if (auto c = l.x <=> r.x; !std::is_eq(c)) /* 1) */ return c;
        if (auto c = l.y <=> r.y; !std::is_eq(c)) return c;
        return l.y <=> r.y;
    }
    // == is not automatically defined in terms of <=>.
    friend constexpr bool operator==(const S&, const S&) = default;
};

If all members of S weren't the same type, then we could either specify the category explicitly (in the return type), or we could obtain it with std::common_comparison_category:

std::common_comparison_category_t<decltype(l.x <=> l.x), /* ... */>

1) Helper functions like std::is_neq compare the result of <=> to zero. They express intent more clearly, but you don't have to use them.

Common Idioms

Alternatively, we can let std::tie figure out the details:

#include <tuple>

struct S {
    int x, y, z;

    friend constexpr auto operator<=>(const S& l, const S& r) noexcept {
        return std::tie(l.x, l.y, l.z) <=> std::tie(r.x, r.y, r.z);
    }
};

Use std::lexicographical_compare_three_way for array members.

Solitaire answered 6/9, 2023 at 20:53 Comment(0)
S
2

Summary of Canonical Function Signatures

Many operator overloads can return pretty much anything. For example, nothing stops you from returning void in operator==. However, only a few of these signatures are canonical, which means that you would normally write them that way, and that such an operator can be explicitly defaulted with = default.

Assignment Operators

struct X {
  X& operator=(const X&) = default;     // copy assignment operator
  X& operator=(X&&) noexcept = default; // move assignment operator
};

Explicit defaulting with = default; is possible, but you can also implement assignment manually. Move assignment is almost always noexcept, although it isn't mandatory.

Comparison Operators

#include <compare> // for comparison categories

struct X {
  friend auto operator<=>(const X&, const X&) = default; // defaulted three-way comparison
  friend std::strong_ordering<=>(const X&, const X&);    // manual three-way comparison

  friend bool operator==(const X&, const X&) = default;  // equality comparisons
  friend bool operator!=(const X&, const X&) = default;  // defaultable since C++20

  friend bool operator<(const X&, const X&) = default;   // relational comparisons
  friend bool operator>(const X&, const X&) = default;   // defaultable since C++20
  friend bool operator<=(const X&, const X&) = default;
  friend bool operator>=(const X&, const X&) = default;
};

See this answer for more information on when and how to default/implement comparisons.

Arithmetic Operators

struct X {
  friend X operator+(const X&, const X&); // binary plus
  friend X operator*(const X&, const X&); // binary multiplication
  friend X operator-(const X&, const X&); // binary minus
  friend X operator/(const X&, const X&); // binary division
  friend X operator%(const X&, const X&); // binary remainder

  X operator+() const;                    // unary plus
  X operator-() const;                    // unary minus

  X& operator++();                        // prefix increment
  X& operator--();                        // prefix decrement
  X  operator++(int);                     // postfix increment
  X  operator--(int);                     // postfix decrement

  X& operator+=(const X&);                // compound arithmetic assignment
  X& operator-=(const X&);
  X& operator*(const X&);
  X& operator/=(const X&);
  X& operator%=(const X&);
};

It is also possible to take the left operator of binary operators by value, but this is not recommended because it makes the signature asymmetric and inhibits compiler optimizations.

Bitwise Operators

struct X {
  using difference_type = /* some integer type */;

  friend X operator&(const X&, const X&);         // bitwise AND
  friend X operator|(const X&, const X&);         // bitwise OR
  friend X operator^(const X&, const X&);         // bitwise XOR
  
  friend X operator<<(const X&, difference_type); // bitwise left-shift
  friend X operator>>(const X&, difference_type); // bitwise right-shift

  X operator~() const;                            // bitwise NOT

  X& operator&=(const X&);                        // compound bitwise assignment
  X& operator|=(const X&);
  X& operator^(const X&);
  X& operator/=(const X&);
  X& operator%=(const X&);
};

Stream Insertion and Extraction

#include <ostream> // std::ostream
#include <istream> // std::istream

struct X {
  friend std::ostream& operator<<(std::ostream&, const X&); // stream insertion
  friend std::istream& operator>>(std::istream&, X&);       // stream extraction
};

Function Call Operator

struct X {
  using result = /* ... */;

         result operator()(user-defined-args...) /* const / volatile / & / && */;
  static result operator()(user-defined-args...);           // since C++23
};

Subscript Operator

struct X {
  using key_type = /* ... */;
  using value_type = /* ... */;

  const value_type& operator[](key_type) const;
        value_type& operator[](key_type);

  static value_type& operator[](key_type); // since C++23
};

Note that operator[] can accept multiple parameters since C++23.

Member Access Operators

struct X {
  using value_type = /* ... */;

  const value_type& operator*() const; // indirection operator
        value_type& operator*();

  const value_type* operator->() const; // arrow operator
        value_type* operator->();
};

Pointer-to-Member Operator

struct X {
  using member_type = /* ... */;
  using member_pointer_type = /* ... */;
  
  const member_type& operator->*(member_pointer_type) const;
        member_type& operator->*(member_pointer_type);
};

Address-of Operator

struct X {
  using address_type = /* ... */;

  address_type operator&() const; // address-of operator
};

Logical Operators

struct X {
  friend X operator&&(const X&, const X&); // logical AND
  friend X operator||(const X&, const X&); // logical OR
  friend X operator!(const X&);            // logical NOT
};

Note that these don't return bool because they only make sense if X is already a logical type that similar to bool.

User-defined Conversions

struct X {
  using type = /* ... */;

  operator type() const;          // arbitrary implicit conversion

  explicit operator bool() const; // explicit/contextual conversion to bool

  template <typename T>
    requires /* ... */            // optionally constrained
  explicit operator T() const;    // conversion function template
};

Coroutine Await

struct X {
  using awaiter = /* ... */;

  awaiter operator co_await() const;
};

Comma Operator

struct X {
  using pair_type = /* ... */;

  // often a template to support combination of arbitrary types
  friend pair_type operator,(const X&, const X&);
};

Allocation Functions

struct X {
  // class-specific allocation functions
  void* operator new(std::size_t);
  void* operator new[](std::size_t);
  void* operator new(std::size_t, std::align_val_t); // C++17
  void* operator new[](std::size_t, std::align_val_t); // C++17

  // class-specific placement allocation functions
  void* operator new(std::size_t, user-defined-args...);
  void* operator new[](std::size_t, user-defined-args...);
  void* operator new(std::size_t, std::align_val_t, user-defined-args...); // C++17
  void* operator new[](std::size_t, std::align_val_t, user-defined-args...); // C++17

  // class-specific usual deallocation functions
  void operator delete(void*);
  void operator delete[](void*);
  void operator delete(void*, std::align_val_t); // C++17
  void operator delete[](void*, std::align_val_t); // C++17
  void operator delete(void*, std::size_t);
  void operator delete[](void*, std::size_t);
  void operator delete(void*, std::size_t, std::align_val_t); // C++17
  void operator delete[](void*, std::size_t, std::align_val_t); // C++17

  // class-specific placement deallocation functions
  void operator delete(void*, user-defined-args...);
  void operator delete(void*, user-defined-args...);

  // class-specific usual destroying deallocation functions
  void operator delete(X*, std::destroying_delete_t); // C++20
  void operator delete(X*, std::destroying_delete_t, std::align_val_t); // C++20
  void operator delete(X*, std::destroying_delete_t, std::size_t); // C++20
  void operator delete(X*, std::destroying_delete_t, std::size_t, std::align_val_t); // C++20
};

// non-class specific replaceable allocation functions ...

void* operator new(std::size_t);
void* operator delete(void*);
// ...
Solitaire answered 15/9, 2023 at 13:9 Comment(1)
This, too, is a very helpful addition. Thank you!Kati
P
-3

Making it short and simple, I'll be referring to some points, which I had come over the past week as I was learning Python and C++, OOP and other things, so it goes as follows:

  1. The arity of the operator can not be modified further than to what it is!

  2. Overloaded operators can only have one default argument which the function call operator rest it cannot.

  3. Only built-in operators can be overloaded, and the rest can't!

For more information, you can refer to Rules for operator overloading, which redirects you to the documentation provided by GeeksforGeeks.

Pound answered 13/3, 2022 at 12:40 Comment(0)

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