C：Typedef结构和指针异常

Question

I was on my Data Structures classes and this confused me.

It's related with pointers properties i guess, but on my research I didn't find any real explanation, any idea why C allows this?

Run-able code: http://ideone.com/kgh3LF

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

/* Declaring a typedef struct  */
typedef struct{
    int a;
    char b[10];
}struct_one;

/* Declaring another structure, with an intentional wrong calling of the first structure */
struct struct_two{
    int p;
    char q[10];

    /* This doesn't work as expected... should be: struct_one var; */
    // struct struct_one var;

    /* THIS ONE DOES WORK!!, and i'm not sure why */
    struct struct_one *ptr;
};

int main(void) {
    /* code */
    return 0;
}

Answer 1

You incorrectly assumed that your second structure "calls" the first one. It doesn't.

In reality the first struct declaration declares as untagged struct type with typedef alias struct_one . The only way to refer to this type is struct_one . Just struct_one , not struct struct_one .

The second struct declaration declares struct struct_two , which refers to a type struct struct_one . This latter type has absolutely nothing to do with the previously declared type struct_one : struct struct_one and struct_one are two completely different, unrelated types. Your reference to struct struct_one inside the second struct is seen as an introduction , a declaration of a completely new type struct struct_one . The compiler assumes that you will define that type later (if necessary).

Since you are not doing anything else in your code, you do not run into any problems caused by such mistake. As an additional illustration of what could happen, see the following example

int main(void) {
    struct_one *p = 0;
    struct struct_two s2;
    s2.ptr = p;       
    return 0;
}

This code will immediately produce a diagnostic message from the compiler, because s2.ptr = p; assignment is illegal: the pointer types are unrelated. The pointer of the left hand side is struct struct_one * and the pointer on the right-hand side is just struct_one * .

Once again, in C language when you use struct <something> syntax in contexts that do not require a complete type, you can write virtually anything in place of <something> : a complete gibberish (as long as it is a lexically valid identifier). If the type is not yet known, the compiler will simply assume that you are introducing a new type, eg

int main() {
  struct klshcjkzdcdsamcbsj78q43698 *p = 0;
}

The above is a perfectly valid C program.

Answer 2

This is allowed because sometimes you have two struct types that each contain a pointer to the other. You need to be able to declare a pointer to the second type in the first one, even though the second struct hasn't been declared yet. If you couldn't forward reference a struct pointer type, you'd have the paradox that each one has to be declared before the other.

struct first {
    int a;
    struct second *second_ptr; // This is allowed
};
struct second {
    int b;
    struct first *first_ptr;
}