Question

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Can a local variable's memory be accessed outside its scope?

I recently came across the following code:

#include <stdio.h>

int* abc () {
   int a[3] = {1,10,100};
   return a;
}
int* xyz () {
   int b[1] = {222};
   return b;
}
int main() {
   int *a, *b;
   a = abc();
   b = xyz();
   printf("%d\n", *a);
   return 0;
}

the output is 222. 'a' is pointing to the array declared inside the xyz().

my question is:

1. why is a pointing to the array declared inside xyz().

2. the array declared inside the function xyz() should go out of scope after the execution of the function. why is that not happening ?

Answer 1

You might see 222 because the memory that was used for the local array in abc has been used for something else – the stack for the function xyz. And you’re passing around an address to that memory. Make a few more function calls and *a may contain some other value.

should go out of scope after the execution of the function. why is that not happening ?

The variable has gone out of scope. Using that address outside the function is incorrect code: using a pointer to local data returned from a function is undefined behavior.