Reference to variable out-of-scope

Question

I have made the following example to test my understanding of references:

#include <iostream>

int test(){
    int a = 1;
    int &b = a;
    return b;
}

int main(int argc, const char * argv[]) {
   
    std::cout << test() << std::endl;
}

What I intended to do is to write an example where the referenced variable is destroyed. I originally wrote this thinking that, since the lifetime of a in test ends when returning b in test, outputting the return-value of test produces gibberish. However, to my big surprise, it actually does output 1!

How is this possible? What am I missing here - can a somehow continue living in b?

Answer 1

The function does not return a reference to any object. It returns a temporary object of the type int that was copy-initialized by the value referenced by b.

int test(){
    int a = 1;
    int &b = a;
    return b;
}

In fact the function is equivalent to the following function without using intermediate variables

int test(){
    return 1;
}

To return a reference means that the function return type should be a referenced type. Something like

int & test()
{
    int a = 1;
    return a;
}

In this case the compiler will issue a message that the function returns a reference to a local variable.