What's the difference between const int& and int&

Question

I got 3 questions.

1.what is the difference between the flowing 2 codes?

//why no errors during compiling as flowing
const int& get3() {
    return 3;
}

and

//get a compiler error, says: 'return': cannot convert from 'int' to 'int &'
//and that's understandable
int& get3() {
    return 3;
}

2.Why the first one is OK to compile?

3.And when i run the first one, i got the strange result:

error image

That's why?

I'll appreciate it very much if anyone can give me some tips.

Answer 1

With int& get3() you can do, for example, get3() = 5.

This will set the value of the variable whose reference is returned by function get3 to 5.

Since 3 is a constant value with no address to be used as reference, you get a compilation error.

Such functions typically return a reference to a class member variable.

Answer 2

2. The first one is actually not OK, even though it compiles. It returns a reference to a local temporary which is undefined behavior (because that temporary gets destroyed after function exits, so you get a dangling reference). The compiler should have already told you about this by issuing a warning.

   Basically it's equivalent to the following:

   const int& get3() {
       const int& x = 3; // bind const reference to a temporary, OK
       return x; // return a reference to a temporary, UB
   }
   

   And this also answers your third question:

3. Because you have UB in your code.

I'd also recommend to read this brilliant answer.