C Two structs cause segmentation fault 11

Question

I'm trying to run this code. If it is ran like this it works, but as soon as the second struct is uncommented it fails with a "Segmentation fault: 11". What am I doing wrong here?

#include <stdio.h>

void func();

typedef struct foo
{
    double one;
    double two;
} foo;


int main() {

    func();
    printf("ret");

    foo *f;
    f->one = 10;
    f->two = 10;

//    foo *g;
//    g->one = 10;
//    g->two = 10;


return 0;
}

You're dereferencing an uninitialized pointer. There's also an undefined function `func` so you should get a link error. — interjay, Oct 17 '17 at 13:23
With `foo *f;` you define a pointer variable `f`. But *where does it point*? Remember that local variables in C are not initialized when defined, their values are *indeterminate* (and seemingly random). — Some programmer dude, Oct 17 '17 at 13:23
@Ôrel Technically the variable `f` itself is allocated. However, there's no allocated memory for it to point to. — Some programmer dude, Oct 17 '17 at 13:24
Not looking at the compiler warnings. That's what you are doing wrong. — n. m. could be an AI, Oct 17 '17 at 13:27
Compile with -Wall -Wextra. This is undefined behavior which means ANYTHING may happen, including the program working as it is supposed to. — klutt, Oct 17 '17 at 13:28

score 3 · Accepted Answer · answered Oct 17 '17 at 13:38

3

You probably intended to do this:

foo f;
f.one = 10;
f.two = 10;

Here f is not a pointer to a foo but it is a foo.

When you write foo *f;, then f is a pointer, and you need to assign a valid memory address to it before you can dereference it, for example as in Marievi's answer.

You should probably read the chapter dealing with pointers in your C text book.

answered Oct 17 '17 at 13:38

Jabberwocky

48,281
17
65
115

wow I did not realise that this was the mistake. I have had a rather brief introduction to C, I'll take another look at the chapter about pointers. – förschter Oct 17 '17 at 15:09

score 2 · Answer 2 · answered Oct 17 '17 at 13:29

2

You declare pointer foo but you also need to allocate memory for it :

foo *f = malloc(sizeof(struct foo));
if (f == NULL)
    return;

because right now, you have it uninitialized but try to de-reference it.

answered Oct 17 '17 at 13:29

Marievi

4,951
1
16
33

CIsForCookies · Answer 3 · 2017-10-17T13:53:15.390

UB warning!!!

Here - foo *g; you only declare a pointer. It points to memory you don't yet own (haven't malloc() it). Trying to de-reference such a pointer is UB and can cause (in your case does cause, a segfault).

Note that the same goes for your foo *f;. malloc() both and them.

Oh, and READ COMPILER WARNINGS!! (warning: 'g' is used uninitialized in this function)

Two ways to fix this:

don't use pointers... change foo *g; into foo g and g->one = 10 into g.one = 10. in this case, the memory is allocated to you on declaration
allocate memory... add this line g = malloc(sizeof(*g)); after foo *g; (and also after foo *f). _here, the memory is allocated by you and would have to be released afterwards by you!! (using free(g); )

C Two structs cause segmentation fault 11

3 Answers3