Pointers to local var - unclear effect

[SeBeQ]

Hi everyone I typed example to learn:

...

void replaceNumberAndPrint(int *x) {
    printf("%d\n", *x);
}

int * getInt() {
    int x = 5;
    return &x;
}


int _tmain(int argc, _TCHAR* argv[])
{
	replaceNumberAndPrint(getInt()); // -858993460 Ok, because point to local var (random piece of memory)
	
	printf("%d\n",*getInt()); // 5 - is good because value have 5 but it pointed to local variable
	
return 0;
}
...

Why first replaceNumberAndPrint(getInt()); show some piece of memory,
and getInt() show local variable value? In my mind second must have value likewise first? Sorry for my English

[Fransje]

The memory that was first used by x and that is set free when getInt() returns, still has the value of 5. Only if another program uses that memory, it will be overwritten. So sometimes *getInt() will give you the correct value (but never rely on it), and sometimes garbage.

[SeBeQ]

Thanks, I understand it now.

If compiler will be faster I have code like this:

- return localNumber
- compiler delete localNumber
- function get address from garbage
- print trash/rubbish

but is a bit slower
- return localNumber
- function get address from localNumber
- print localNumber
- compiler delete localNumber

[Cubbi]

The behavior of that program is undefined (because it attempts to dereference a pointer to a variable that no longer exists).
I'll run it through a few more compilers for fun (after replacing _tmain with main of course)
gcc/linux:

0
0

gcc/ibm:

512547305
0

xlc/ibm:

5
5

sun:

-71350040
-71350040

All these compilers are behaving correctly