C ++ implicit conversion constructor call

Question

Why does gcc require a copy constructor to invoke an implicit conversion constructor?

class X { public: X(int q) {} ~X() { std::cout << "~X()" << std::endl; } X(const X&) = delete; }; X x = 1; // gives error: use of deleted function 'X::X(const X&)' 

More interesting here, even if I write a copy constructor, it is not called. The destructor is called only once, so the following code

 class X { public: X(int q) {} ~X() { std::cout << "~X()" << std::endl; } X(const X&) { std::cout << "copy ctor" << std::endl; } }; int main() { X x = 1; } 

Fingerprints ~X()

This is mistake? Is there any workaround?

The gcc version on my locaL machine is 4.6.3, this also works on a different version of gcc (online)

http://ideone.com/ustDRj

Answer 1

X x = 1; is the syntax for copy initialization. If the initializer does not have type X (as here), this is semantically equivalent to this:

 X x(X(1)); 

That is, it creates ins from argument 1 , and then copies-initializes X from this instance.

Like any other initialization of a copy, the copy can be deleted. This probably happens in your case, so the copy constructor is not actually called. However (again, as with any other instance), it should be available.

Answer 2

you are trying to initialize a value for a class object here that invokes a copy constructor.

 int main() { X x = 1; } 

change the code to

 X x(1);