Changing value after calling variable by using pointers in C

Question

so I started programming in C. Now I have a problem with pointers:

int * diff(){
  int array[2] = {1,2};
  return array;
}

int main(int argc, char const *argv[]) {
  int *p;
  p = diff();
  printf("%d\n", *(p));
  printf("%d\n", *(p));

  return 0;
}

So after starting the program. My terminal is showing the following:

1
0

So why is the Value of *p changing ?

Returning a pointer to local variable == undefined behavior. — Fred Larson, Apr 06 '17 at 18:41
@Downvoters, a little harsh given the question is well-written and a good code snippet. As a beginner it's difficult to conjure up the search terms. — Bathsheba, Apr 06 '17 at 18:43

score 1 · Accepted Answer · answered Apr 06 '17 at 18:41

1

The behaviour of your program is undefined.

array has automatic storage duration (informally, think of this as a "local variable"), and dereferencing the pointer to it that's returned back to main is not allowed by the language.

(We call this a dangling pointer).

answered Apr 06 '17 at 18:41

Bathsheba

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Sir, nitpick: _"dereferencing the pointer to it that's returned back to main is not allowed by the language"_, well, the language never stops you, all it says that it'll be UB. – Sourav Ghosh Apr 06 '17 at 18:45

Changing value after calling variable by using pointers in C

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