warning: address of local variable ‘f’ returned

Code:

#include<stdio.h>

struct face
{
  int c;                                                                                                                                                    
};
typedef struct face face;

face * get_face()
{
  face f = {4};
  return &f;
}

int main()
{                                                                                                                        
  face * f = get_face();
  printf("face c: %d\n", f->c);
                                                                                                                                                                                                                                         
  return 0;
}

Here is the warning I get:

warning: address of local variable ‘f’ returned


Since I'm creating face on the stack, shouldn't it go away once the function exits? However, it only throws a warning and still runs perfectly fine (prints 4).

Why does this run correctly? Is it guaranteed to run correctly?

It runs correctly because you were lucky. It is not guaranteed to run correctly.

The memory formerly known as f is still there, however it is no longer reserved by the stack and can be overwritten at any point in time. The common misunderstanding is that the memory gets filled with 0s or something when the variables are "removed" from the stack. All that is happening is the memory is no longer reserved.

It does "go away". The memory that was formerly occupied by it is now free to be overwritten by anything else that follows. Even how the printf call itself on the very next line down doesn't happen to overwrite that particular piece of memory before displaying the value is just pure dumb luck.

It's just as though you were in a takeaway shop where they give out a number when you order, and then call out that number when they've made your meal. Holding onto that number for the following day doesn't mean that when they call that number out again that the food is for you. It might be, if you re-ordered and happened to be given the same number, but it quite likely is not.
So don't go holding onto memory addresses of things that are no longer valid.

C code should really go into the C section, not the C++ section.

Originally Posted by Elysia

C code should really go into the C section, not the C++ section.

It is pretty hard for a beginner to distinguish C code from C++ code, yet when it compiles on both compilers.

A beginner should know what language they are studying.

@OP:
Try this code

Code:

#include<stdio.h>

int* get_local_var(int i)
{
	int ret = i;
	return &ret;
}

int main()
{
	int* i1 = get_local_var(1);
	int* i2 = get_local_var(2);
	printf("i1: %d\n", *i1);

	return 0;
}

i1 should print 1, right?

@Others:
Yes I know this code relies on undefined behaviour and it might not work as I intended. But I'm assuming that since the OP's code worked for him, this will to. I'm just trying to prove a point as to why you shouldn't keep pointers to local variables across function boundaries.

Originally Posted by _Mike

@OP:
Try this code

Code:

#include<stdio.h>

int* get_local_var(int i)
{
	int ret = i;
	return &ret;
}

int main()
{
	int* i1 = get_local_var(1);
	int* i2 = get_local_var(2);
	printf("i1: %d\n", *i1);

	return 0;
}

i1 should print 1, right?

ret is a local variable, you can't pass its address because it's gone when get_local goes out of scope. If you want to return local variable, return by value.

Code:

int get_local_var(int i)
{
	int ret = i;
     // do somethign else to ret here
	return ret;
}

Unless you allocate the memory on the heap, then yes, you can return the pointer. But that is a pretty bad practice becuase you will forget to free it later.

Code:

int * get_local_var(int i)
{
  int* ret=malloc(sizeof (int));
  *p=i;
  return p; <-------- have to free this in main() or whereever. When you have a hundred of these pointers, it becomes impossible to keep track of them. 
}

Originally Posted by nimitzhunter

ret is a local variable, you can't pass its address because it's gone when get_local goes out of scope. If you want to return local variable, return by value.

Originally Posted by _Mike

@Others: Yes I know this code relies on undefined behaviour and it might not work as I intended. But I'm assuming that since the OP's code worked for him, this will to. I'm just trying to prove a point as to why you shouldn't keep pointers to local variables across function boundaries.