Structures and pointers in C

Question

I'm creating a structure inside a function and then returning its pointer. For some reason I keep getting the same memory address each time.

typedef struct list_type
{
    node_t *head;
    node_t *tail;
} list_t;

list_t newList() {
    list_t list = {NULL, NULL};
    list_t *listptr = &list;
    printf("newList: %p\n", listptr);
    return listptr;
}

Outputs:

newList: 0x7fffb42c8ae0
newList: 0x7fffb42c8ae0
newList: 0x7fffb42c8ae0

What am I doing?...

Read about [undefined behavior](http://en.wikipedia.org/wiki/Undefined_behavior). You got one. — Basile Starynkevitch, Sep 25 '14 at 05:41
For one, `list_t*` isn't `list_t`. The compiler should have at-least warned you of this. If it didn't, ram up your warning level. — WhozCraig, Sep 25 '14 at 05:42
This code shouldn't compile as you return `list_t*` while have `list_t` in definition. — Геннадий Казачёк, Sep 25 '14 at 05:42
Looks like you come from a javascript or python background -- C does not work the same way -- you need to explicitly allocate your structures if you want to use them the way you do. — Soren, Sep 25 '14 at 06:43
@BasileStarynkevitch The code he has contains no undefined behavior. OP has not shown any code that actually uses the pointer returned (which could be undefined behavior). — fuz, Sep 25 '14 at 11:29

score 3 · Answer 1 · answered Sep 25 '14 at 05:40

You are allocating list on the stack, but you're trying to return a reference to it from a function. This is a big no-no in C, since the struct's memory is automatically freed by the compiler when it goes out of scope. If you want memory to persist beyond the scope it is allocated in, you need to use malloc.

Furthermore, list_t and a pointer to it (list_t*) are two different types, and you need to indicate that as such in your function definition.

list_t* newList() {
    list_t *listptr = malloc(sizeof(list_t));
    printf("newList: %p\n", listptr);
    return listptr;
}

However, be careful when using malloc. Since you're allocating memory manually, you'll now need to ensure that you call free on the pointer when you're done with it, or it will leak, since the compiler cannot free the memory for you.

Neha Gangwar · Answer 2 · 2014-09-25T06:57:29.523

3

list_t list = {NULL, NULL}; allocates memory on stack. Memory allocated on stack is automatically freed when the variable, list in your case, goes out of scope. That is why you are getting the same address again and again as memory allocated to list on stack gets freed when program comes out of function newList() and is available for allocation. And, when you call the newList() again, same memory is alloacted again.

You should read about different memory allocations and pay more attention to the compiler warnings.

edited Sep 25 '14 at 06:57

answered Sep 25 '14 at 06:00

Neha Gangwar

670
9
14

+1 for pay more attention to the compiler warnings.. – Hemant Gangwar Sep 25 '14 at 06:17

score 2 · Answer 3 · answered Sep 25 '14 at 05:40

2

You're returning a pointer to an object allocated on the stack. If you use the value that the pointer points to in the function that calls newList(), you'll cause undefined behavior to occur.

You should read about malloc() and free().

answered Sep 25 '14 at 05:40

Bill Lynch

80,138
16
128
173

WhozCraig · Answer 4 · 2014-09-25T05:53:03.173

Your function returns a list_t, a structure type, which is perfectly fine by-value. I see no evidence you're trying to dynamically allocate a list_t, and no evidence you even need to do so.

Lose the pointer stuff for this specific structure type entirely and just do this:

list_t newList() 
{
    list_t list = {NULL, NULL};
    return list;
}

void freeList(list_t lst)
{
    // TODO: delete lst nodes by walking lst.head through lst.tail
}

int main()
{
    list_t lst = newList();

    ...

    freeList(lst);
}

Don't step into the arena of managing dynamic memory because you can; do it when you need to do so, and in this case, you don't. You obviously need to in order to manage the actual list nodes, but thats a different issue than this one.

Structures and pointers in C

4 Answers4