Question

File api.h

#include <stdio.h>
#ifndef API
#define API

struct trytag;
typedef struct trytag try;

void trial (try *);

#endif

File core.h

#ifndef CORE
#define CORE
struct trytag
{
    int a;
    int b;
};
#endif

File func.c

#include "api.h"
#include "core.h"

void trial (try *tryvar)
{
    tryvar->a = 1;
    tryvar->b = 2;
}

File main.c

#include "api.h"

int main ()
{
    try s_tryvar;

    trial(&s_tryvar);

    printf("a = %d\nb = %d\n", s_tryvar.a, s_tryvar.b);
}

When I compile, I get:

main.c:5: error: storage size of ‘s_tryvar’ isn’t known

If I include core.h in main.c this error doesn’t come as try is defined in core.h. But I want the structure try to be hidden to main.c — it should not know the members of try structure. What am I missing?

Answer 1

I don’t think what you’re trying to do is possible. The compiler needs to know how big a try structure is to compile main.c. If you really want it to be opaque, make a generic pointer type, and instead of declaring the variable directly in main(), make alloc_try() and free_try() functions to handle the creation and deletion.

Something like this:

api.h:

#ifndef API
#define API

struct trytag;
typedef struct trytag try;

try *alloc_try(void);
void free_try(try *);
int try_a(try *);
int try_b(try *);
void trial (try *);

#endif

core.h:

#ifndef CORE
#define CORE
struct trytag
{
    int a;
    int b;
};
#endif

func.c:

#include "api.h"
#include "core.h"
#include <stdlib.h>

try *alloc_try(void)
{
    return malloc(sizeof(struct trytag));
}

void free_try(try *t)
{
    free(t);
}

int try_a(try *t)
{
    return t->a;
}

int try_b(try *t)
{
    return t->b;
}

void trial(try *t)
{
    t->a = 1;
    t->b = 2;
}

main.c:

#include <stdio.h>
#include "api.h"

int main()
{
    try *s_tryvar = alloc_try();

    trial(s_tryvar);
    printf("a = %d\nb = %d\n", try_a(s_tryvar), try_b(s_tryvar));

    free_try(s_tryvar);
}