C - forward declare and use struct not pointer

Question

I'm trying to forward declare struct_A and use it within struct_B not as a pointer but as a structure. Within struct_B, I make reference to struct_B.

I can't seem to get it right - help? Here's my code:

struct type_A_t;
typedef struct type_A_t type_A_t;

typedef struct type_B_t {
    type_A_t a;
};

struct  type_A_t{
    void (*cb)(type_B_t *b1);
    void (*cb)(type_B_t *b2);
};

The error I get is:

error: field 'type_A_t' has incomplete type

another problem is that type_B_t is not defined

M.M
– M.M

2018-12-18 09:19:31 +00:00
Commented Dec 18, 2018 at 9:19 — M.M
– M.M, Commented Dec 18, 2018 at 9:19

Blaze · Accepted Answer · 2018-12-18 09:19:34Z

3

You can't do that. But in your example, since type_A_t only needs to know about pointers to type_B_t, why not switch it around like this?

typedef struct type_B_t type_B_t;
typedef struct type_A_t type_A_t;

struct  type_A_t {
    void(*cb)(type_B_t *b1);
    void(*cb)(type_B_t *b2);
};
typedef struct type_B_t {
    type_A_t a;
};

answered Dec 18, 2018 at 9:19

Blaze

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Angew is no longer proud of SO · Accepted Answer · 2018-12-18 09:20:30Z

0

That is not possible. To use a type as a member, the type's size must be known, so it must be fully defined. Otherwise, sizeof(struct type_B_t) would be unknown after its definition, which makes no sense.

answered Dec 18, 2018 at 9:20

Angew is no longer proud of SO

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gsamaras · Accepted Answer · 2018-12-18 09:21:51Z

0

I can't seem to get it right - help?

Because it cannot be done.

The compiler doesn't know the size of that struct.

A forward declaration is enough for pointers though:

typedef struct type_B_t {
    type_A_t a;  // error
    type_A_t* a; // OK
};

Of course, if in your real code, you can switch the order, as @Blaze's answer mention, then do that.

edited Dec 18, 2018 at 9:21

answered Dec 18, 2018 at 9:18

gsamaras

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1 Comment

Angew is no longer proud of SO Over a year ago

This is a C only question, there are no references.

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