C global pointer value changes after irrelevant function call

Question

I run the following code

int *pointer;

void fun1() {
    int i;
    pointer = &i;
    *pointer = 11;
}

void fun2() {
    printf("\nFun 2: Hello World\n");
}

int main() {
    pointer = (int*) malloc(sizeof(int));
    *pointer = 0;
    fun1();
    printf("%d\n",*pointer);
    fun2();
    printf("%d\n",*pointer);
    return 0;
}

And the output is

11

Fun 2: Hello World
20

I would expect it to be 11 instead of 20. What am I missing?

Thanks.

You are taking a pointer to a local variable. `i` get's invalid/random after your function returns. Possible duplicate to [Why does gcc throw a warning when returning a pointer to a local variable and not when returning a local variable?](http://stackoverflow.com/questions/8184315/why-does-gcc-throw-a-warning-when-returning-a-pointer-to-a-local-variable-and-no) and [returning a local variable from function in C](http://stackoverflow.com/questions/4824342/returning-a-local-variable-from-function-in-c) — Florian, Jun 23 '15 at 17:56

haccks · Answer 1 · 2015-06-23T18:00:32.427

1

This an example of undefined behavior. When fun1 is called, pointer is pointing to an automatic local variable in fun1. Once fun1 completed its execution that variable no longer exist. Now the statement

printf("%d\n",*pointer);

invoke undefined behavior.
There is also memory leak in your program.

edited Jun 23 '15 at 18:00

answered Jun 23 '15 at 17:57

haccks

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Then how the first print gives 11 ? fun1 has ended at this point as well, shouldn't it be undefined behavior as well? – kopsti Jun 23 '15 at 17:58
*Then how the first print gives 11 ?*: Compiler is joking! – haccks Jun 23 '15 at 18:01
Isn't it what Wormbo says? – kopsti Jun 23 '15 at 18:03
@illeg; Well its UB and anything could happen. – haccks Jun 23 '15 at 18:06

score 0 · Accepted Answer · answered Jun 23 '15 at 17:58

0

You set the pointer to local variable i's address in fun1 and it stays that way after fun1 returns. That address is on the stack and later is overwritten with a new value during your call to fun2 and its printf call.

answered Jun 23 '15 at 17:58

Wormbo

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mathematician1975 · Answer 3 · 2015-06-23T18:04:47.190

0

Here in fun1()

pointer = &i;

you assign the global pointer to point to a temporary variable that no longer exists when fun1() exits. Therefore when you call printf with *pointer you are accessing invalid memory.

This is undefined behaviour and trying to understand the content of the invalid memory region afterwards is not worthwile.

edited Jun 23 '15 at 18:04

answered Jun 23 '15 at 17:58

mathematician1975

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C global pointer value changes after irrelevant function call

3 Answers3