Question

I do not understand how this code is compiling. Can somebody please explain what is going on in there.

#include <iostream>

using namespace std;

class B
{
public:    
    B(const char* str = "\0") //default constructor
    {
        cout << "Constructor called" << endl;
    }    

    B(const B &b)  //copy constructor
    {
        cout << "Copy constructor called" << endl;
    } 
};

int main()
{ 
    B ob = "copy me";    //why no compilation error.
    return 0;
}

The optput is:
Constructor called

P.S.: I could not think of a more apt title than this, Anyone who can think of a better title, please modify it.

Answer 1

The type of "copy me" is char const[8], which decays to char const *. Since the default constructor is not explicit, "copy me" can be implicitly converted to B, and thus ob can be copy-constructed from that implicitly converted, temporary B-object.

Had the default constructor been declared explicit, you would have had to write one of the following:

B ob1 = B("copy me");
B ob2("copy me");

Had the copy constructor also been declared explicit, you would have had to say one of these:

B ob3(B("copy me"));
B ob4("copy me");

In practice, all copies will be elided by any half-decent compiler, and you always end up with a single default constructor invocation.