在复制初始化中，对复制构造函数的调用是显式还是隐式的？

Question

class AAA {
public:
    explicit AAA(const AAA&) {}
    AAA(int) {}
};


int main() {
    AAA a = 1;
    return 0;
}

In the above code, as I understand, though elided in most cases, the copy constructor is still semantically required to be called. My question is, is the call explicit or implicit? For a long time I have the conclusion in my mind that the call to AAA::AAA(int) is implicit but the call to the copy constructor is not. Today I accidentally got g++ to compile the above code and it reported error. (VC12 compiles OK.)

In section 8.5 of the standard:

If the destination type is a (possibly cv-qualified) class type:

• If the initialization is direct-initialization, or if it is copy-initialization where the cv-unqualified version of the source type is the same class as, or a derived class of, the class of the destination, constructors are considered. The applicable constructors are enumerated (13.3.1.3), and the best one is chosen through overload resolution (13.3). The constructor so selected is called to initialize the object, with the initializer expression or expression-list as its argument(s). If no constructor applies, or the overload resolution is ambiguous, the initialization is ill-formed.

• Otherwise (ie, for the remaining copy-initialization cases), user-defined conversion sequences that can convert from the source type to the destination type or (when a conversion function is used) to a derived class thereof are enumerated as described in 13.3.1.4, and the best one is chosen through overload resolution (13.3). If the conversion cannot be done or is ambiguous, the initialization is ill-formed. The function selected is called with the initializer expression as its argument; if the function is a constructor, the call initializes a temporary of the cv-unqualified version of the destination type. The temporary is a prvalue. The result of the call (which is the temporary for the constructor case) is then used to direct-initialize , according to the rules above, the object that is the destination of the copy-initialization. In certain cases, an implementation is permitted to eliminate the copying inherent in this direct-initialization by constructing the intermediate result directly into the object being initialized; see 12.2, 12.8.

The bolded direct-initialize in the above quotes means the call to copy constructor is explicit, right? Is g++ wrong or my interpretation of the standard wrong?

Answer 1

Looks like this bug: g++ fails to call explicit constructors in the second step of copy initialization

g++ fails to compile the following code

 struct X { X(int) {} explicit X(X const &) {} }; int main() { X x = 1; // error: no matching function for call to 'X::X(X)' } 

The second step of a copy initialization (see 8.5/16/6/2) is a direct-initialization where explicit constructors shall be considered as candidate functions.

Answer 2

Looks like copy constructor is never called . Only constructor is called . The following code may call copy constructor

AAA a = 1;
AAA ab = a;

Not sure why G++ is compiling it .