C Declare the struct before definition

Question

The following is the source code that is problem:

#include <stdio.h>

typedef struct Foo
{
    int a;
    Bar b;
} Foo;

typedef struct Bar
{
    int a;
    int b;
} Bar;

int main(void)
{
    Foo f;
    f.a = 1;
    f.b.a = 2;
    f.b.b = 3;
    printf("%d %d %d\n", f.a, f.b.a, f.b.b);

    return 0;
}

I want to declare the Bar struct in the Foo struct that keep order of the struct declarations (Foo struct is first). But I can't, it occurs some errors(One is error: unknown type name 'Bar') in the compile time.

How to do it?

Answer 1

The compiler needs to be able to determine the size of Foo. If Bar is unknown at the time Foo is defined the compiler can not determine the size of Foo. The only way around this is using a pointer since all pointers have the same size.

You can use a forward declaration of the structure and then reference it as a pointer. This means that Foo can never automatically allocate the memory for Bar. As a consequence the memory has to be allocated separately.

If you can avoid this do not do it.

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

typedef struct Bar Bar;
typedef struct Foo Foo;

struct Foo
{
    int a;
    Bar * b;
};

struct Bar
{
    int a;
    int b;
};

void dynamic(void)
{
    Foo f;

    f.a = 1;
    f.b = (Bar*)malloc(sizeof(Bar));
    f.b->a = 2;
    f.b->b = 3;
    printf("%d %d %d\n", f.a, f.b->a, f.b->b);

    free(f.b);
}

void automatic(void)
{
    Foo f;
    Bar b;

    f.a = 1;
    f.b = &b;
    f.b->a = 2;
    f.b->b = 3;
    printf("%d %d %d\n", f.a, f.b->a, f.b->b);
}

int main(void)
{
    dynamic();
    automatic();
}