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

13.3.3.2/3 says

Two implicit conversion sequences of the same form are indistinguishable conversion sequences unless one
of the following rules apply:

— Standard conversion sequence S1 is a better conversion sequence than standard conversion sequence
S2 if :

S1 and S2 are reference bindings (8.5.3), and the types to which the references refer are the same
type except for top-level cv-qualifiers, and the type to which the reference initialized by S2 refers is
more cv-qualified than the type to which the reference initialized by S1 refers. [Example:

int f(const int &);
int f(int &);
...
int i;
int j = f(i); // calls f(int&)

In your case since the argument is non-const, the non-const version of the copy c-tor is chosen because it is a better match.