Question

Consider the following code:

#include <iostream>
using namespace std;
class B{
public:
  B(){}
};
class A
{
public:
A(){}
A(B &b){

}
A(const B &b){
       cout<<"cccddd"<<endl;
}
};
int main()
{
   B b;
   A c(b);
   A a;
   a=b; //ok
   A &ref = b; //error and why???
}

why b assigned to a is ok,but b assigned to ref is illegal???

Answer 1

A(const B &b){
       cout<<"cccddd"<<endl;
}

This will enable you to convert an object of type B into an object of type A. However, a reference T& cannot be created from another type L.

8.5.3 References [dcl.init.ref]

1. A variable declared to be a T& or T&&, that is, “reference to type T” (8.3.2), shall be initialized by an object, or function, of type T or by an object that can be converted into a T.

2. […]

3. […]

4. Given types “cv1 T1” and “cv2 T2,” “cv1 T1” is reference-related to “cv2 T2” if T1 is the same type as T2, or T1 is a base class of T2. “cv1 T1” is reference-compatible with “cv2 T2” if T1 is reference-related to T2 and cv1 is the same cv-qualification as, or greater cv-qualification than, cv2 […].

In your example neither are A and B the same type, nor is A a base class of B, thus both classes aren’t reference-related.

Why won’t A &ref = (A)b work?

The expression (A) b will actually create a temporary, which cannot be bound to a normal reference:

error: invalid initialization of non-const reference of type ‘A&’ from a temporary of type ‘B’

You would need to a const reference A const & ref = (A)b for this.