default constructor call vs. copy constructor call

Question

if I create three objects like this:

A myA; // line 1
A myA2 = A(); // line 2
A myA3 = myA; // line 3

I thought in the second line the copy constructor is called. But if I try this (Visual Studio) line 1 and line 2 calls only the default constructor. Whereas of course line 3 calls the copy constructor. So it seems there is no difference between line 1 and line 2. Both calls the default construcor. So why I have the different syntax?

By the way my testclass A has the form:

class A { 
  public:
   A() { cout << "default ctor is called..." << endl; }
   A(const A &obj) { cout << "copy ctor is called..." << endl; }
};

See: [What are copy elision and return-value optimization?](https://stackoverflow.com/questions/12953127/what-are-copy-elision-and-return-value-optimization). — WhozCraig, May 11 '17 at 08:52

score 0 · Answer 1 · answered May 11 '17 at 08:53

The copy constructor calling is optimized according to copy elision; which is guaranteed since C++17.

In initialization, if the initializer expression is a prvalue and the cv-unqualified version of the source type is the same class as the class of the destination, the initializer expression is used to initialize the destination object

default constructor call vs. copy constructor call

1 Answers1