Can you declare an anonymous instance of a named struct?

I'm trying to manually implement a polymorphic behavior in C by creating a generic struct, and then derived structs (if you will) which can be told apart by the value of an enum, so that I can have a pointer to the generic type, dereference it as the generic type, figure out what type it is, and then dereference it as the more specific type.

typedef struct{
  enum type structType;
  //... other stuff all the structs share
}generic;

typedef struct{
  generic; //does not work, obviously, nor does (generic){};
  //... other stuff unique to struct type A
}typeA;

I understand that I could just declare a named instance of the generic struct in the derived struct, but this seems a little messy, and I would prefer not to if there's a neat and tidy way around it.

You can't always get what you want, but if you try sometimes, well, you might find, you get what you need ...

There are two basic ways, with a slight bit of trickery:

1. Using an include file (e.g.): generic.h
2. Using a CPP macro (e.g): GENERIC

I've used both methods at various times.

---

Here's the method with the include file (generic.h):

enum type structType;
int com_fd;
void *com_buf;

And, here's a .c file that uses it:

typedef struct {
#include <generic.h>
} generic;

typedef struct {
#include <generic.h>
    // other stuff unique to struct type A ...
    int typea_value;
} typeA;

---

Here's the method using a macro:

#define GENERIC \
    enum type structType; \
    int com_fd; \
    void *com_buf

typedef struct {
    GENERIC;
} generic;

typedef struct {
    GENERIC;

    // other stuff unique to struct type A ...
    int typea_value;
} typeA;