What does the explicit keyword mean?

Question

What does the explicit keyword mean in C++?

Conchiolin

I just want to point out to anyone new coming along that ever since C++11, explicit can be applied to more than just constructors. It's now valid when applied to conversion operators as well. Say you have a class BigInt with a conversion operator to int and an explicit conversion operator to std::string for whatever reason. You'll be able to say int i = myBigInt;, but you'll have to cast explicitly (using static_cast, preferably) in order to say std::string s = myBigInt;. – Bodice

@Bodice There is an explicit keyword that can be used on the declaration of an implicit conversion. – Mitis

@curiousguy: What do you mean? All conversions should be implicit? To let loose all sorts of silently applied funny miscasts due to accidental ambiguities? (See e.g. the "The safe bool problem" section at this C++ref. page, or open-std.org/jtc1/sc22/wg21/docs/papers/2007/n2333.html for (much) more details on why "Explicit conversion is an ill defined concept." is an ill-conceived statement.) – Holleran

@Holleran I mean that explicit conversion is not a thing; it's a garbage concept. Also the "safe bool" is a ridiculous "proof" for the usefulness of the "explicit operator" idea as it isn't even an application of that idea but a different set of rules, which means the only practical use of "explicit operator" in the SL is ad hoc and not applicable to UDT. – Mitis

@Bodice did you mean like this: std::string s = static_cast<std::string>(myBigInt) ? If possible, could you please elaborate furthermore on your first comment? Thanks a lot in advance! – Nub

@Milan, Yes, that's exactly it. If you're looking for more information, this answer writes it up more formally. Do note that bool is special in this regard. Those answers and searching "explicit conversion operators" will lead you to more writeups about this feature and be better suited than a comment chain. – Bodice

Answer 1

The compiler is allowed to make one implicit conversion to resolve the parameters to a function. This means that the compiler can use constructors callable with a single parameter to convert from one type to another in order to get the right type for a parameter.

Here's an example with converting constructors that shows how it works:

struct Foo {
    // Single parameter constructor, can be used as an implicit conversion.
    // Such a constructor is called "converting constructor".
    Foo(int x) {}
};
struct Faz {
    // Also a converting constructor.
    Faz(Foo foo) {}
};

// The parameter is of type Foo, not of type int, so it looks like
// we have to pass a Foo.
void bar(Foo foo);

int main() {
    // However, the converting constructor allows us to pass an int.
    bar(42);
    // Also allowed thanks to the converting constructor.
    Foo foo = 42;
    // Error! This would require two conversions (int -> Foo -> Faz).
    Faz faz = 42;
}

Prefixing the explicit keyword to the constructor prevents the compiler from using that constructor for implicit conversions. Adding it to the above class will create a compiler error at the function call bar(42). It is now necessary to call for conversion explicitly with bar(Foo(42))

The reason you might want to do this is to avoid accidental construction that can hide bugs.
Contrived example:

• You have a MyString class with a constructor that constructs a string of the given size. You have a function print(const MyString&) (as well as an overload print (char *string)), and you call print(3) (when you actually intended to call print("3")). You expect it to print "3", but it prints an empty string of length 3 instead.

Conchiolin

nice write up, you might want to mention multi-arg ctors with default params can also act as single arg ctor, e.g., Object( const char* name=NULL, int otype=0). – Bischoff

I think it should also be mentioned that one should consider making single argument constructors explicit initially (more or less automatically), and removing the explicit keyword only when the implicit conversion is wanted by design. I think contructors should be explicit by default with an 'implicit' keyword to enable them to work as implicit conversions. But that's not how it is. – Edulcorate

@thecoshman: You don't declare a parameter explicit -- you declare a constructor explicit. But yes: your parameters of type Foo have to be constructed explicitely, they won't be silently constructed by just plugging their constructor's parameters into the function. – Capelin

Just an FYI that when calling "print(3)" in your example, the function needs to be "print(const MyString &"). The "const" is mandatory here because 3 is converted to a temporary "MyString" object and you can't bind a temporary to a reference unless it's "const" (yet another in a long list of C++ gotchas) – Sisson

For completeness sake, I am adding that in addition to parameter conversion the explicit keyword here will also prevent the use of assignment form of a copy ctor (e.g., Foo myFoo = 42;) and require the explicit forms Foo myFoo = Foo(42); or Foo myFoo(42); – Townsfolk

But, if I understand your latter counter-example correctly, print("3") should yield a compiling error since the function print is actually expecting a MyString instance, and not an actual string, and the constructor of the MyString class does not allow the compiler to automatically cast the string "3" to a MyString object. Am I understanding it wrong? – Childbirth

Answer 2

Suppose, you have a class String:

class String {
public:
    String(int n); // allocate n bytes to the String object
    String(const char *p); // initializes object with char *p
};

Now, if you try:

String mystring = 'x';

The character 'x' will be implicitly converted to int and then the String(int) constructor will be called. But, this is not what the user might have intended. So, to prevent such conditions, we shall define the constructor as explicit:

class String {
public:
    explicit String (int n); //allocate n bytes
    String(const char *p); // initialize sobject with string p
};

Cyr

And it's worth noting that the new generalized initialization rules of C++0x will make String s = {0}; ill-formed, rather than trying to call the other constructor with a null pointer, as String s = 0; would do. – Skylar

Even though this is an old question it seems worth pointing a few things out (or having someone set me straight). By making the int form, or both ctors, 'explicit' you would still have the same bug if you used String mystring('x') when you meant String mystring("x") wouldn't you? Also, from the comment above I see the improved behavior of String s = {0} over String s = 0 thanks to making the int form of the ctor 'explicit'. But, other than knowing the precedence of the ctors how do you know the intent (i.e. how to spot the bug) of this String s{0} ? – Townsfolk

Why String mystring = 'x'; is getting converted to int? – Consultation

@InQusitive: 'x'is being treated as an integer because the char data type is just a 1-byte integer. – Excurved

The problem with your example is that it only works with copy initialization (using =) but not with direct initialization (without using =): the compiler will still call the String(int) constructor without generating an error if you write String mystring('x');, as @Townsfolk pointed out. The explicit keyword is meant for preventing implicit conversions that happen in direct initialization and function resolution. A better solution to your example would be a simple overload of the constructor: String(char c);. – Outdo

I am confused. String mystring = 'x' would still invoke the same constructor with an int type even with an explicit keyword no? – Antoninaantonino

Answer 3

In C++, a constructor with only one required parameter is considered an implicit conversion function. It converts the parameter type to the class type. Whether this is a good thing or not depends on the semantics of the constructor.

For example, if you have a string class with constructor String(const char* s), that's probably exactly what you want. You can pass a const char* to a function expecting a String, and the compiler will automatically construct a temporary String object for you.

On the other hand, if you have a buffer class whose constructor Buffer(int size) takes the size of the buffer in bytes, you probably don't want the compiler to quietly turn ints into Buffers. To prevent that, you declare the constructor with the explicit keyword:

class Buffer { explicit Buffer(int size); ... }

That way,

void useBuffer(Buffer& buf);
useBuffer(4);

becomes a compile-time error. If you want to pass a temporary Buffer object, you have to do so explicitly:

useBuffer(Buffer(4));

In summary, if your single-parameter constructor converts the parameter into an object of your class, you probably don't want to use the explicit keyword. But if you have a constructor that simply happens to take a single parameter, you should declare it as explicit to prevent the compiler from surprising you with unexpected conversions.

Christian

useBuffer expects an lvalue for his argument, useBuffer(Buffer(4)) will not work either because of it. Changing it to take a const Buffer& or Buffer&& or just Buffer would make it work. – Oscillation

Answer 4

The keyword explicit accompanies either

• a constructor of class X that cannot be used to implicitly convert the first (any only) parameter to type X

C++ [class.conv.ctor]

1) A constructor declared without the function-specifier explicit specifies a conversion from the types of its parameters to the type of its class. Such a constructor is called a converting constructor.

2) An explicit constructor constructs objects just like non-explicit constructors, but does so only where the direct-initialization syntax (8.5) or where casts (5.2.9, 5.4) are explicitly used. A default constructor may be an explicit constructor; such a constructor will be used to perform default-initialization or valueinitialization (8.5).

• or a conversion function that is only considered for direct initialization and explicit conversion.

C++ [class.conv.fct]

2) A conversion function may be explicit (7.1.2), in which case it is only considered as a user-defined conversion for direct-initialization (8.5). Otherwise, user-defined conversions are not restricted to use in assignments and initializations.

Overview

Explicit conversion functions and constructors can only be used for explicit conversions (direct initialization or explicit cast operation) while non-explicit constructors and conversion functions can be used for implicit as well as explicit conversions.

/*
                                 explicit conversion          implicit conversion

 explicit constructor                    yes                          no

 constructor                             yes                          yes

 explicit conversion function            yes                          no

 conversion function                     yes                          yes

*/

Example using structures X, Y, Z and functions foo, bar, baz:

Let's look at a small setup of structures and functions to see the difference between explicit and non-explicit conversions.

struct Z { };

struct X { 
  explicit X(int a); // X can be constructed from int explicitly
  explicit operator Z (); // X can be converted to Z explicitly
};

struct Y{
  Y(int a); // int can be implicitly converted to Y
  operator Z (); // Y can be implicitly converted to Z
};

void foo(X x) { }
void bar(Y y) { }
void baz(Z z) { }

Examples regarding constructor:

Conversion of a function argument:

foo(2);                     // error: no implicit conversion int to X possible
foo(X(2));                  // OK: direct initialization: explicit conversion
foo(static_cast<X>(2));     // OK: explicit conversion

bar(2);                     // OK: implicit conversion via Y(int) 
bar(Y(2));                  // OK: direct initialization
bar(static_cast<Y>(2));     // OK: explicit conversion

Object initialization:

X x2 = 2;                   // error: no implicit conversion int to X possible
X x3(2);                    // OK: direct initialization
X x4 = X(2);                // OK: direct initialization
X x5 = static_cast<X>(2);   // OK: explicit conversion 

Y y2 = 2;                   // OK: implicit conversion via Y(int)
Y y3(2);                    // OK: direct initialization
Y y4 = Y(2);                // OK: direct initialization
Y y5 = static_cast<Y>(2);   // OK: explicit conversion

Examples regarding conversion functions:

X x1{ 0 };
Y y1{ 0 };

Conversion of a function argument:

baz(x1);                    // error: X not implicitly convertible to Z
baz(Z(x1));                 // OK: explicit initialization
baz(static_cast<Z>(x1));    // OK: explicit conversion

baz(y1);                    // OK: implicit conversion via Y::operator Z()
baz(Z(y1));                 // OK: direct initialization
baz(static_cast<Z>(y1));    // OK: explicit conversion

Object initialization:

Z z1 = x1;                  // error: X not implicitly convertible to Z
Z z2(x1);                   // OK: explicit initialization
Z z3 = Z(x1);               // OK: explicit initialization
Z z4 = static_cast<Z>(x1);  // OK: explicit conversion

Z z1 = y1;                  // OK: implicit conversion via Y::operator Z()
Z z2(y1);                   // OK: direct initialization
Z z3 = Z(y1);               // OK: direct initialization
Z z4 = static_cast<Z>(y1);  // OK: explicit conversion

Why use explicit conversion functions or constructors?

Conversion constructors and non-explicit conversion functions may introduce ambiguity.

Consider a structure V, convertible to int, a structure U implicitly constructible from V and a function f overloaded for U and bool respectively.

struct V {
  operator bool() const { return true; }
};

struct U { U(V) { } };

void f(U) { }
void f(bool) {  }

A call to f is ambiguous if passing an object of type V.

V x;
f(x);  // error: call of overloaded 'f(V&)' is ambiguous

The compiler does not know wether to use the constructor of U or the conversion function to convert the V object into a type for passing to f.

If either the constructor of U or the conversion function of V would be explicit, there would be no ambiguity since only the non-explicit conversion would be considered. If both are explicit the call to f using an object of type V would have to be done using an explicit conversion or cast operation.

Conversion constructors and non-explicit conversion functions may lead to unexpected behaviour.

Consider a function printing some vector:

void print_intvector(std::vector<int> const &v) { for (int x : v) std::cout << x << '\n'; }

If the size-constructor of the vector would not be explicit it would be possible to call the function like this:

print_intvector(3);

What would one expect from such a call? One line containing 3 or three lines containing 0? (Where the second one is what happens.)

Using the explicit keyword in a class interface enforces the user of the interface to be explicit about a desired conversion.

As Bjarne Stroustrup puts it (in "The C++ Programming Language", 4th Ed., 35.2.1, pp. 1011) on the question why std::duration cannot be implicitly constructed from a plain number:

If you know what you mean, be explicit about it.

Lewison

"If you know what you mean be explicit about it." What a great quote. Remind me why auto exists again? :-) – Teratism

@Teratism auto exists to make C++ easier to write and harder to read :( – Matron

that's great until you're the guy who inherits the system. – Teratism

Answer 5

This answer is about object creation with/without an explicit constructor since it is not covered in the other answers.

Consider the following class without an explicit constructor:

class Foo
{
public:
    Foo(int x) : m_x(x)
    {
    }

private:
    int m_x;
};

Objects of class Foo can be created in 2 ways:

Foo bar1(10);

Foo bar2 = 20;

Depending upon the implementation, the second manner of instantiating class Foo may be confusing, or not what the programmer intended. Prefixing the explicit keyword to the constructor would generate a compiler error at Foo bar2 = 20;.

It is usually good practice to declare single-argument constructors as explicit, unless your implementation specifically prohibits it.

Note also that constructors with

• default arguments for all parameters, or
• default arguments for the second parameter onwards

can both be used as single-argument constructors. So you may want to make these also explicit.

An example when you would deliberately not want to make your single-argument constructor explicit is if you're creating a functor (look at the 'add_x' struct declared in this answer). In such a case, creating an object as add_x add30 = 30; would probably make sense.

Here is a good write-up on explicit constructors.

Adoration

Answer 6

The explicit keyword makes a conversion constructor to non-conversion constructor. As a result, the code is less error prone.

Pampas

Answer 7

The explicit-keyword can be used to enforce a constructor to be called explicitly.

class C {
public:
    explicit C() =default;
};

int main() {
    C c;
    return 0;
}

the explicit-keyword in front of the constructor C() tells the compiler that only explicit call to this constructor is allowed.

The explicit-keyword can also be used in user-defined type cast operators:

class C{
public:
    explicit inline operator bool() const {
        return true;
    }
};

int main() {
    C c;
    bool b = static_cast<bool>(c);
    return 0;
}

Here, explicit-keyword enforces only explicit casts to be valid, so bool b = c; would be an invalid cast in this case. In situations like these explicit-keyword can help programmer to avoid implicit, unintended casts. This usage has been standardized in C++11.

Polyamide

explicit operator bool() is also the C++11 version of safe bool, and can be used implicitly in condition checks (and only in condition checks, as far as I'm aware). In your second example, this line would also be valid in main(): if (c) { std::cout << "'c' is valid." << std:: endl; }. Apart from this, though, it can't be used without explicit casting. – Vibrate

"constructor to be called explicitly" no – Mitis

@JustinTime It's an inane, broken version of the safe bool. The whole idea of explicit implicit conversion is absurd. – Mitis

@Mitis True. It seems a bit like a kludge, aimed more at being easily remembered (likely in the hopes of that translating to frequently used) than at following English logic, and designed to not be outright incompatible with previous safe bool implementations (so you're less likely to break something if you swap it in). IMO, at least. – Vibrate

I edited this to fix the C c(); and also played around in Compiler Explorer and it appears that making a default constructor explicit doesn't actually do anything. This answer is simply wrong, even with the code corrections already made. – Macadam

Answer 8

Cpp Reference is always helpful!!! Details about explicit specifier can be found here. You may need to look at implicit conversions and copy-initialization too.

Quick look

The explicit specifier specifies that a constructor or conversion function (since C++11) doesn't allow implicit conversions or copy-initialization.

Example as follows:

struct A
{
    A(int) { }      // converting constructor
    A(int, int) { } // converting constructor (C++11)
    operator bool() const { return true; }
};

struct B
{
    explicit B(int) { }
    explicit B(int, int) { }
    explicit operator bool() const { return true; }
};

int main()
{
    A a1 = 1;      // OK: copy-initialization selects A::A(int)
    A a2(2);       // OK: direct-initialization selects A::A(int)
    A a3 {4, 5};   // OK: direct-list-initialization selects A::A(int, int)
    A a4 = {4, 5}; // OK: copy-list-initialization selects A::A(int, int)
    A a5 = (A)1;   // OK: explicit cast performs static_cast
    if (a1) cout << "true" << endl; // OK: A::operator bool()
    bool na1 = a1; // OK: copy-initialization selects A::operator bool()
    bool na2 = static_cast<bool>(a1); // OK: static_cast performs direct-initialization

//  B b1 = 1;      // error: copy-initialization does not consider B::B(int)
    B b2(2);       // OK: direct-initialization selects B::B(int)
    B b3 {4, 5};   // OK: direct-list-initialization selects B::B(int, int)
//  B b4 = {4, 5}; // error: copy-list-initialization does not consider B::B(int,int)
    B b5 = (B)1;   // OK: explicit cast performs static_cast
    if (b5) cout << "true" << endl; // OK: B::operator bool()
//  bool nb1 = b2; // error: copy-initialization does not consider B::operator bool()
    bool nb2 = static_cast<bool>(b2); // OK: static_cast performs direct-initialization
}

Flagon

explicit operator bool() vs. if is a special case. There is no way to reproduce it with user defined Bool, explicit operator Bool() and a function called If. – Mitis

Answer 9

It is always a good coding practice to make your one argument constructors (including those with default values for arg2,arg3,...) as already stated. Like always with C++: if you don't - you'll wish you did...

Another good practice for classes is to make copy construction and assignment private (a.k.a. disable it) unless you really need to implement it. This avoids having eventual copies of pointers when using the methods that C++ will create for you by default. An other way to do this is derive from boost::noncopyable.

Slum

This post is written in 2009. Today you don't declare them as private, but rather say = delete. – Quilmes

Is the word 'explicit' missing in your first sentence? Should it read, "explicit as already stated"? – Claudette

Answer 10

Constructors append implicit conversion. To suppress this implicit conversion it is required to declare a constructor with a parameter explicit.

In C++11 you can also specify an "operator type()" with such keyword http://en.cppreference.com/w/cpp/language/explicit With such specification you can use operator in terms of explicit conversions, and direct initialization of object.

P.S. When using transformations defined BY USER (via constructors and type conversion operator) it is allowed only one level of implicit conversions used. But you can combine this conversions with other language conversions

• up integral ranks (char to int, float to double);
• standart conversions (int to double);
• convert pointers of objects to base class and to void*;

Softshoe

Answer 11

Other answers are missing one important factor which I am going to mention here.

Along with "delete" keyword, "explicit" allows you to control the way compiler is going to generate special member functions - default constructor, copy constructor, copy-assignment operator, destructor, move constructor and move-assignment.

Refer https://learn.microsoft.com/en-us/cpp/cpp/explicitly-defaulted-and-deleted-functions

Jdavie

This is not even an answer. And more, delete keyword does NOT allow you to control the way that the compiler is going to generate special member functions. – Temporary