Why the first copy constructor is called in the code below ?

Question

Why the B(B&) ctor is called, instead of B(const B&), in the construction of object b1 ?

#include <iostream>
using namespace std;

struct B
{
    int i;
    B() : i(2) { }
    B(B& x) : i(x.i) { cout << "Copy constructor B(B&), i = " << i << endl; }
    B(const B& x) : i(x.i) { cout << "Copy constructor B(const B&), i = " << i << endl; }
};

int main()
{
    B b;
    B b1(b);
}

Answer 1

This is because overload resolution applies, and since the argument to the constructor of b1 is b, and b happens to be non-const lvalue, then the constructor taking non-const lvlalue is selected. And that's the first one. Interestingly, both are copy constructors, but your code would be equaly valid with just the latter one.

Answer 2

Because b is not const. Therefore, it matches the first copy ctor perfectly, so that's what the compiler uses.