Array inside struct filled with garbage after initialization

Question

I defined a data structure (Foo) that contains an array of 25 pointers (members) to itself. I want to initialize each of those pointers to NULL, but my init function isn't working correctly. When my Foo f returns from foo_init(), only some of the members are NULL, others are just populated with random values.

//in Foo.h
#include <stdio.h>
#include <stdlib.h>

typedef struct Foo {
    struct Foo * members[25] ;
} Foo ;

void foo_init(Foo * f) ;

//in Foo.c
void foo_init(Foo * f) {
    f = (Foo*)malloc(sizeof(Foo));

    for (size_t i = 0 ; i < 25 ; i++) {
        f->members[i] = NULL ;
    }
    /* Ok here, all members are NULL */
}

//in main.c
#include "Foo.h"

int main(int argc, const char * argv[])
{

    Foo f ;
    foo_init(&f) ;

    /* why isn't every index of f.members NULL? */


    /* ... */
    return 0;
}

I ran my code through LLDB. Inside foo_init(), all members are NULL. But after returning from foo_init(), f.members is full of random garbage values.

You are allocating heap memory for something that was already declared on the stack in your first line in foo_init. Get rid of that line. — wolfPack88
– wolfPack88, Commented Jun 20, 2014 at 12:07

diapir · Accepted Answer · 2014-06-20 12:11:45Z

5

The address of the Foo in your main is immediately overwritten by the call to malloc. Just remove that line and your initialization should work.

If you really want foo_init to allocate memory you can convert it to :

void foo_init(Foo **out_f) {
    Foo *f = malloc(sizeof *f);

    for (size_t i = 0 ; i < 25 ; i++) {
        f->members[i] = NULL;
    }

    *out_f = f;
}

Or simply return a pointer as it is idiomatic in C :

Foo *foo_init(void);

edited Jun 20, 2014 at 12:11

answered Jun 20, 2014 at 12:06

diapir

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2 Comments

Orelsanpls Over a year ago

Did the compilator know the type of *f in moment of sizeof? Thx i learned something today :)

Aravind A Over a year ago

Can you please tell me why do you use a pointer to a pointer Foo **out_f?

Orelsanpls · Accepted Answer · 2014-06-20 12:07:42Z

2

Try this then

Foo *f ;
foo_init(&f) ;


void foo_init(Foo ** f) {
    *f = (Foo*)malloc(sizeof(Foo));

    for (size_t i = 0 ; i < 25 ; i++) {
        (*f)->members[i] = NULL ;
    }
}

@TripeHound is right

edited Jun 20, 2014 at 12:07

answered Jun 20, 2014 at 12:02

Orelsanpls

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2 Comments

AdamJames Over a year ago

Whoops, that was just a typo. My actual code already looks like Foo f ; foo_init(&f) ;

TripeHound Over a year ago

Which won't work because the first thing the function does is throw away the passed-in value and malloc a new structure.

Serve Laurijssen · Accepted Answer · 2014-06-20 12:14:43Z

1

another option is to use a pointer to pointer if you want to allocate Foo inside foo_init(). This way f points to valid memory after returning from foo_init

void foo_init(Foo ** f) {
    *f = (Foo*)malloc(sizeof(Foo));

    for (size_t i = 0 ; i < 25 ; i++) {
        (*f)->members[i] = NULL ;
    }
}

int main()
{
    Foo *f;

    foo_init(&f);

    return 0;
}

answered Jun 20, 2014 at 12:14

Serve Laurijssen

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