c++: cast operator vs. assign operator vs. conversion constructor priority

Question

Let's have this code:

Test1 t1;
Test2 t2;
t1 = t2;

I believe there are three (or more?) ways how to implement t1 = t2

• to overload assign operator in Test1
• to overload type cast operator in Test2
• to create Test1(const Test2&) conversion constructor

According to my GCC testing, this is the priority of what is used:

1. assign operator
2. conversion constructor and type cast operator (ambiguous)
3. const conversion constructor and const type cast operator (ambiguous)

Please help me understand why this priority.

I use this code for testing (uncomment some lines to try out)

struct Test2;
struct Test1 {
  Test1() { }
  Test1(const Test2& t) { puts("const constructor wins"); }
//  Test1(Test2& t) { puts("constructor wins"); }
//  Test1& operator=(Test2& t) { puts("assign wins"); }
};

struct Test2 {
  Test2() { }
//  operator Test1() const { puts("const cast wins"); return Test1(); }
//  operator Test1() { puts("cast wins"); return Test1(); }
};


int main() {
  Test1 t1;
  Test2 t2;
  t1 = t2;
  return 0;
}

Tabloid

Test1::Test1(const Test2&) is not a "copy constructor", it is a "converting constructor". – Occasion

This post explains exactly why conversion operator is of higher precedence: #1384507 – Whichever

Answer 1

The statement t1 = t2; is equivalent to:

t1.operator=(t2);

Now the usual rules of overload resolution apply. If there's a direct match, that's the chosen one. If not, then implicit conversions are considered for use with the (automatically generated, "implicitly defined") copy-assignment operator.

There are two possible implicit, user-defined conversions. All user-defined conversions count equal, and if both are defined, the overload is ambiguous:

• Convert t2 to a Test1 via the Test1::Test1(Test2 const &) conversion constructor.

• Convert t2 to a Test1 via the Test2::operator Test1() const cast operator.

Turves

In @Luchian's second ideone example, the conversion function wins because it binds a Test2& to t2, not a const Test2&. I thought ideone.com/U38vK was supposed to be ambiguous too, but it seems g++ prefers the constructor. – Occasion

Aha. g++ DOES call them ambiguous if you ask for -pedantic. Naughty default g++. – Occasion

@KerrekSB: If the cast operator is not constant, it beats the conversion constructor and there is no ambiguity. liveworkspace.org/code/7795254ae49b4d6350f0ede57615e4c6 – Stagnate

@JanTuroň: Yes, I know. And if you make the constructor non-const as well, you can get the ambiguity back. – Turves

@JanTuroň: The exact same reason. – Turves

@KerrekSB: OK, why the non-constant version of operator/constructor wins over the constant version of constructor/operator? – Stagnate

@JanTuroň: Because t2 is non-const. You can also try things like t1 = static_cast<Test2 const &>(t2); and compare. – Turves

@KerrekSB: OK, now it is clear. Thank you, it was a long struggle. – Stagnate

@JanTuroň: Awesome. Overload resolution is a very complex part of C++, so it's important to understand it right. – Turves

@KerrekSB : since when does it make any sense for a constructor to be const or non-const ? Constructors cannot be cv-qualified – Segarra

Answer 2

when I use following code than priority is given first to the constructor rather than cast operator

 #include<iostream>
 using namespace std;
 class C1;
 class C2
 {
      int x;
 public:
     operator C2()
     {
       C2 temp;
       cout<<"operator function called"<<endl;
       return temp;
    }
 };
class C1
{
   int x;
public:
   C1():x(10){}
   C1(C2)
  {
    cout<<"constructor called"<<endl;
  }
};
 int main()
{
  C1 obj1;
  C2 obj2;
  obj1=obj2;
}

Output Constructor called

Pigfish