C - function returning a pointer to a local variable

Question

Consider the following code.

#include<stdio.h>
int *abc(); // this function returns a pointer of type int

int main()
{
    int *ptr;
    ptr = abc();
    printf("%d", *ptr);
    return 0;
}

int *abc()
{
    int i = 45500, *p;
    p = &i;
    return p;
}

Output:

45500

I know according to link this type of behavior is undefined. But why i am getting correct value everytime i run the program.

Because undefined behaviour is allowed to do that. By the way, I don't see anything wrong with this question. Upvoted. — Bathsheba
– Bathsheba, Commented Sep 7, 2016 at 15:11
Its because your program is very simple. Try to insert a call to some other function between calls to abc() and printf() — Serge
– Serge, Commented Sep 7, 2016 at 15:11
Add this, immediately after the call to abc() and before the existing printf: printf("Test string %d %d %d %d\n", 1, 2, 3, 4). Then see what happens :) — kdopen
– kdopen, Commented Sep 7, 2016 at 15:20
Call abc() then call def() and after check what p points to... You could make i static, as long as there is no reentrance issue... (e.g. threads modifying *p ...) — Déjà vu
– Déjà vu, Commented Sep 7, 2016 at 15:34

redneb · Accepted Answer · 2016-09-07 15:23:22Z

6

Every time you call abc it "marks" a region at the top of the stack as the place where it will write all of its local variables. It does that by moving the pointer that indicates where the top of stack is. That region is called the stack frame. When the function returns, it indicates that it does not want to use that region anymore by moving the stack pointer to where it was originally. As a result, if you call other functions afterwards, they will reuse that region of the stack for their own purposes. But in your case, you haven't called any other functions yet. So that region of the stack is left in the same state.

All the above explain the behavior of your code. It is not necessary that all C compilers implement functions that way and therefore you should not rely on that behavior.

edited Sep 7, 2016 at 15:23

answered Sep 7, 2016 at 15:15

redneb

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17 Comments

Eugene Sh. Over a year ago

This is very implementation specific.

redneb Over a year ago

Yes, I am only explaining the behavior to satisfy his curiosity.

Serge Over a year ago

As an example Keil C compiler for i8051: It allocates locals in data segment by building a calling tree at link time, so stack is used only to pass control

user3185968 Over a year ago

@Serge: How does it handle recursion?

kdopen Over a year ago

Which is where the spec meets reality. I've had this discussion with others. Sometimes, there is no way to create a conformant compiler for a given chip architecture. That doesn't mean an almost-compliant compiler isn't better than raw assembler. Also, those compilers are usually C90

|

Sourav Ghosh · Accepted Answer · 2016-09-07 15:14:55Z

2

Well, undefined behavior is, undefined. You can never rely on UB (or on an output of a program invoking UB).

Maybe, just maybe in your environment and for your code, the memory location allocated for the local variable is not reclaimed by the OS and still accessible, but there's no guarantee that it will have the same behavior for any other platform.

edited Sep 7, 2016 at 15:14

answered Sep 7, 2016 at 15:11

Sourav Ghosh

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2 Comments

Cody Over a year ago

what's UB ?? pls define

Sourav Ghosh Over a year ago

@Cody UB -> undefined behavior.

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