C ++函数销毁后会返回有效的“按引用”对象

Question

I'm using visual studio 2010 compiler and I'm trying to understand the output of this program.

Code:

#include <iostream>
using namespace std;

class A 
{
public:
    int i;
    A()
    {
        i=0; cout<<"constructing A..\n";
    }
    A(int a): i(a)
    {
        cout<<"constructing A with argument\n";
    }
    A(A& a)
    {
        i=a.i;
        cout<<"copy constructor\n";
    }
    ~A()
    {
        cout<<"destructing a: " << i << endl;
    }
};

A& f(A b)
{
    return A(25);
}

void main()
{
    A m;
    cout << "i = " << f(m).i << endl;
}

Output:
constructing A..
copy constructor
constructing A with argument
destructing a: 25
destructing a: 0
i = 25
destructing a: 0

From my understanding, A(25) returned by reference and then destroyed, so why does it print the value of i: 'i = 25'?

Answer 1

The program has undefined behavior. Nevertheless the output can be as you expected because the memory occupied by the non alive object can be not overwritten yet.