What is the difference between the outputs of these two programs and why?

Question

Here are two programs. Both of them have a function fun() whose return type is reference to an integer. Only the difference between two functions is that in one function x is declared as static int while in other it is not. The output to the first question is 10 and the output to the second question is 30. How?

Program 1:

#include<iostream>
using namespace std;

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

Program 2:

#include<iostream>
using namespace std;

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

The first program invokes undefined behavior, the second one does not. — , Aug 01 '13 at 19:04
what is undefined behaviour? can u tell me why is the difference in the output of both the programs? — thedarkpassenger, Aug 01 '13 at 19:05
You can google "undefined behavior" and you will understand it. — , Aug 01 '13 at 19:06

score 2 · Answer 1 · answered Aug 01 '13 at 19:08

Program 1: you return a reference to a variable which ceases to exist as soon as the function returns, then store the value 30 to that non-existing variable, which may or may not crash your machine. Regardless, the following call to fun() reinitializes a local variable "x" and returns it.

Program 2: A static variable at function scope is kind of like a global variable that's only accessible to that function. You return a reference to that static (which still exists), and set its value. When you call the function again, the static still has the value you assigned to it.

What is the difference between the outputs of these two programs and why?

1 Answers1