c++ - Does shallow copy call member objects' constructors?

Question

Assuming i have this code:

class A {
public:
    int x;
    A(){}
    A(const A& a){} //copy constructor
    operator= (const A &a){...}
};
class B {
public:
    A a;
    B(){}
};
int main() {
    B b;
    B c = b; //shallow copy
    B d;
    d = b; //shallow assignment
}

Will the shallow copyassignment call member A a's copy constructorassignment operator overloading? Or shortly does shallow copy perform member objects' user-made copy constructor & assignment operator or a shallow one as well?

question from:https://stackoverflow.com/questions/65905010/does-shallow-copy-call-member-objects-constructors

Answer 1

Then the answer is yes, the copy constructor will be called, but the assignment operator will not.

The copy constructor will be called because the defaulted copy constructor is called, but the assignment operator is not. The defaulted assignment operator will call the defaulted copy constructor, and the defaulted copy constructor will call the copy constructor of the base class, but the defaulted assignment operator is not called.

The reason is simple: the defaulted assignment operator is not called because the defaulted assignment operator is declared "A& operator=(const A& a)".

The above is a case in which the calling convention (the "= default" is a member function calling convention) is used.