Outputs changing when using static variables inside function which is returning references in c++

Question

Why does the following yield output as 30 ? while removing "static" from x gives 10 ?

#include<iostream>
using namespace std;

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

Possible duplicate of [Assigning value to function returning reference](https://stackoverflow.com/questions/18964801/assigning-value-to-function-returning-reference) — Gaurav Sehgal, Oct 24 '17 at 04:43
If you remove the `static` you return a reference to a local variable that goes away. Assigning through that reference is undefined behavior, and *anything* can happen. — Bo Persson, Oct 24 '17 at 04:58
@GauravSehgal - that's not quite a duplicate. That question is about how to get the compiler to give warnings in this case, not about why removing `static` gives different behaviour. — Peter, Oct 24 '17 at 07:41

tost · Answer 1 · 2017-10-24T05:57:29.927

using static, 'x' is not stored in the stack, it is stored in the .bss or in .data. So it keeps the value that you assign in the main and returns it when you call fun the second time (that's the meaning of that static).
If you remove static, x is allocated in the stack and when fun() returns it disappears. So the value of x is reinitialized to 10 each time you call fun(). Really, by removing the static, the first line of your main is meaningless because fun() returns a reference to nothing.

score 0 · Answer 2 · answered Oct 24 '17 at 07:38

When x is static, it continues to exist after the function returns. So the assignment fun() = 30 reliably changes the value of x, and cout << fun() reliably prints the last assigned value of x (i.e. 30 in your code).

Without the static keyword, x has automatic storage duration so - as far as your program is concerned - starts to exist when the first statement inside fun() is reached AND ceases to exist when the function returns. The assignment fun() = 30 and the streaming cout << fun() therefore both exhibit undefined behaviour, since they each attempt to use (via a reference) a variable that has ceased to exist.

When behaviour is undefined, anything can happen (since the meaning of "undefined" in the C++ standard is that the C++ standard does not constrain what can happen). In your case, it just so happens a value of 10 (the result of the assignment inside fun()) is printed. That behaviour is not guaranteed - completely different results are also feasible (e.g. results change when using a different compiler, different optimisation settings, or even simply with phase of the moon).

Most compilers can be configured to give a warning when a function returns a reference or pointer to a variable with automatic storage duration. However, most modern compilers are configured by default so they do NOT give such a warning, so you will need to read documentation for your compiler to work out how to enable higher warning levels. If you do that you will find that your compiler will give a warning of some form about fun(), when the static keyword is removed.

Outputs changing when using static variables inside function which is returning references in c++

2 Answers2