为什么在看起来好像不是指针的情况下，此C代码为什么将局部结构视为指针？

Question

In the following code, "stk" is treated as if it is a pointer. But after looking at it from every angle for hours, I cannot for the life of me see how it is a pointer. Can someone please explain what I'm missing?

struct T {
    int count;
    struct elem {
        void *x;
        struct elem *link;
    } *head;
};

T Stack_new(void) {
    T stk;
    NEW(stk);
    stk->count = 0;
    stk->head = NULL;
    return stk;
}

My interpretation says that T is a struct, and therefore stk is a local, automatic variable containing a struct. It is not a pointer, but then it gets treated as a pointer, leaving me stuck in a WTF state.

More Background This code is from a book called "C Interfaces and Implementations" by Hanson. He creates a library of abstract data types that expose an interface and hide the implementation. The stack is the first one he covers. I'm a long-time programmer just now digging into C, and apparently there's some way of parsing this syntax that I'm missing. Thanks.

In case it is relevant, here is the definition for NEW and the things that new calls:

#define  NEW(p) ((p) = ALLOC((long)sizeof *(p)))

#define ALLOC(nbytes) \
    Mem_alloc((nbytes), __FILE__, __LINE__)

extern void *Mem_alloc (long nbytes,
    const char *file, int line);

Answer 1

In the snippet above, T stk will declare stk as a variable of type T . However type T isn't defined anywhere, and the code won't compile.

If it instead said struct T stk; , it would be declaring stk as a variable having type struct T . However, the attempts to dereference stk would be meaningless and the code would again fail to compile.

To make the example work, you could add something like,

typedef struct T *T

which defines type T to be a pointer to struct T . I would find this highly confusing though.