Output in c++ and why?

Question

code

#include<iostream>
using namespace std;


int &fun()
{
    int x = 10;
    return x;
}
int main()
{
    fun() = 30;
    cout << fun();
    return 0;
}

output wil be 10 , tell me how and when int x = 10 is changed to static int x = 10 output will be 30 .Explain both of the cases .

This is phrased almost like an interview question (and if it is, IMO it's not a very good one). — NPE, Jul 21 '14 at 05:40
http://stackoverflow.com/questions/8610350/implications-of-using-an-ampersand-before-a-function-name-in-c — nobalG, Jul 21 '14 at 05:42
@tc: Correct. Output could be: it overwrites your stack and reformats your hard disk. Not a good interview/homework question, unless it's a trick question. — david.pfx, Jul 21 '14 at 06:08
@david.pfx i would say it is a reasonable beginner question. Just not something you should post on SO. — juanchopanza, Jul 21 '14 at 06:15

score 4 · Answer 1 · answered Jul 21 '14 at 05:45

This is undefined behavior. You are returning a reference to a local variable whose lifetime has ended at the end of the function.

It's quite amusing what g++ does with this code:

At -O0, it prints 10.

At -O1, it prints 30.

At -O2 and -O3, it prints 0.

If you declare x as static, then it has static storage duration, which means that its lifetime doesn't end when the function returns, which means that it is legal to return a reference to it. All calls to foo will return a reference to the same int.

Output in c++ and why?

1 Answers1