Allocate space for struct pointer in subfunction

Question

How can I allocate memory for a struct pointer and assign value to it's member in a subfunction?

The following code will compile but not execute:

#include <stdio.h>
#include <stdlib.h>
#include <string.h>

struct _struct {char *str;};
void allocate_and_initialize(struct _struct *s)
{
    s = calloc(sizeof(struct _struct), 1);
    s->str = calloc(sizeof(char), 12);
    strcpy(s->str, "hello world");
}
int main(void)
{
    struct _struct *s;
    allocate_and_initialize(s);
    printf("%s\n", s->str);

    return 0;
}

Answer 1

You are passing s by value. The value of s is unchanged in main after the call to allocate_and_initialize

To fix this you must somehow ensure that the s in main points to the memory chunk allocated by the function. This can be done by passing the address of s to the function:

// s is now pointer to a pointer to struct.
void allocate_and_initialize(struct _struct **s)
{
        *s = calloc(sizeof(struct _struct), 1); 
        (*s)->str = calloc(sizeof(char), 12);
        strcpy((*s)->str, "hello world");                                                                                                                                                                      
}
int main(void)
{
        struct _struct *s = NULL;  // good practice to make it null ptr.
        allocate_and_initialize(&s); // pass address of s.
        printf("%s\n", s->str);

        return 0;
}

Alternatively you can return the address of the chunk allocated in the function back and assign it to s in main as suggested in other answer.

Answer 2

In your example:

void allocate_and_initialize(struct _struct *s)
{
    s = calloc(sizeof(struct _struct), 1);
    s->str = calloc(sizeof(char), 12);
    strcpy(s->str, "hello world");
}

Assigning to s here doesn't change s in the caller. Why not return it instead?

struct _struct *allocate_and_initialize(void) {
    struct _struct *s;
    s = calloc(sizeof *s, 1);
    s->str = calloc(1, 12); /* sizeof(char) is always 1 */
    strcpy(s->str, "hello world");
    return s;
}

and use it thus:

struct _struct *s;
s = allocate_and_initialize();
/* use s... */
free(s); /* don't forget to free the memory when you're done */

Answer 3

you must change your code like that:

    #include <stdio.h>
    #include <stdlib.h>
    #include <string.h>

    struct _struct {char *str;};
    void allocate_and_initialize(struct _struct **s)
    {
        *s = (_struct*)calloc(sizeof(struct _struct), 1);
        (*s)->str = (char*)calloc(sizeof(char), 12);
        strcpy((*s)->str, "hello world");
    }
    int main(void)
    {
        struct _struct *s;
        allocate_and_initialize(&s);
        printf("%s\n", s->str);

        return 0;
    }

The reason is, that you change the adress of the pointer, but not the "content" of the pointer. So, if you code in c, you have to use a "double" pointer. If you code in c++ you can use a reference.