Nameless union inside a union

Question

I'm reading some code and found something like the following:

typedef union {
    int int32;
    int boolean;
    time_t date;
    char *string;
    union {
        struct foo *a;
        struct foo *b;
        struct foo *c;
    };
} type_t;

From syntax point of view, the inner union {} can be removed and having *a, *b and *c directly inside the outer union {}. So what's the purpose the namelessly embedded union?

Answer 1

Unnamed union/struct inside another union/struct is a feature of C11, and some compiler extensions (e.g, GCC).

C11 §6.7.2.1 Structure and union specifiers

13 An unnamed member whose type specifier is a structure specifier with no tag is called an anonymous structure; an unnamed member whose type specifier is a union specifier with no tag is called an anonymous union. The members of an anonymous structure or union are considered to be members of the containing structure or union. This applies recursively if the containing structure or union is also anonymous.

The advantage of this feature is that one can access its unnamed union field easier:

type_t x;

To access the field a, you can simply use x.a. Compare with the code without using this feature:

typedef union {
    int int32;
    int boolean;
    time_t date;
    char *string;
    union u{      //difference in here
    struct foo *a;
    struct foo *b;
    struct foo *c;
    };
} type_t;

type_t x;

You need to use x.u.a.

Related: unnamed struct/union in C