Returning local reference

Question

When I run my code, attempting to return a local variable:

#include<iostream>
using namespace std;

int &fun()
{     
  int x = 30;      
  return x;    
}

int main()    
{   
    fun() = 10;   
    cout << fun();    
    return 0;    
}

why does some compiler output 0 and some are 30

i would say it is very ugly code. what is the use-case? apart from that, you may want to use `static int x = 30;`. — hoijui, Jul 07 '15 at 07:04
@Bathsheba please find more appropriate duplicate then, because there must be some. — Anton Savin, Jul 07 '15 at 07:06

score 3 · Answer 1 · answered Jul 07 '15 at 07:03

3

Returning a reference to a local variable that subsequently goes out of scope is undefined behaviour in C++.

Sometimes it might work, sometimes it might not. Occasionally the compiler might eat your cat.

answered Jul 07 '15 at 07:03

Bathsheba

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I always knew computers ran on kitties and bits. – Jul 07 '15 at 07:14

Returning local reference

1 Answers1