Why is copy-constructor not called in this case?

Question

I encountered a code snippet and thought that it would call copy-constructor but in contrast , it simply called normal constructor . Below is the code

#include <iostream>
using namespace std;
class B
{
    public:    
    B(const char* str = "\0")
    {
        cout << "Constructor called" << endl;
    }    
    B(const B &b)
    {
        cout << "Copy constructor called" << endl;
    } 
};
int main()
{  
    B ob = "copy me"; 
    return 0;
}

Answer 1

What you've discovered that B ob = "copy me"; notionally creates a B from the literal and then copy constructs ob , but that the compiler is allowed to elide the copy and construct directory into ob . g++ even elides the copy with no optimization enabled at all.

You can observe that this is the case by making your copy constructor private: The code will fail to compile even though the compiler won't actually use the copy constructor (the standard requires that copy constructors be accessible even when the call is elided).