why default copy-ctor is generated for a class with reference member variable?

Question

http://coliru.stacked-crooked.com/a/8356f09dff0c9308

#include <iostream>
struct A
{
  A(int& var) : r(var) {}
  int &r;
};

int main(int argc, char** argv)
{
    int x = 23;

    A a1(x);   // why this line is fine?

    A a2 = a1; // why this line is fine?

    a2 = a1; // error: use of deleted function 'A& A::operator=(const A&)'
            // note: 'A& A::operator=(const A&)' is implicitly deleted because the default definition would be ill-formed:
            // error: non-static reference member 'int& A::r', can't use default assignment operator
    return 0;
}

The default assignment operator is deleted. Why the default copy constructor is still kept?

Answer 1

A a1(x);

is fine because it's constructing an instance of A with a reference (x is turned into a reference and the constructor A(int&) is called)

A a2 = a1;

Is also fine because it is still construction. Copy construction, in fact. It's okay to initialize a reference with another reference.

For instance:

int a = 1;
int& b = a;
int& c = b;

Is okay because this is all construction (Demo)

However, you cannot assign a reference, which is what a2 = a1 will attempt to do through a compiler-generated copy-assignment operator. However, the compiler recognized this and did not generate such an operator. Since the operator does not exist, you got a compiler error.