What does (int (*)()) mean in a function call

Question

Like the title says, I want to know what "(int (*)())" in a C-define-function-call means?

As example, it looks similar to this:

#define Bla(x)    (Char *) read((char *(*)()) Blub, (char **) x)

or this

#define XXX(nx, id)     PEM_ASN1_write_bio((int (*)()) id, (char *) nx)

Thank you in advance!

Answer 1

The casts the argument to a pointer to a function that returns char * and takes zero or more arguments. The second function returns int.

You can use a program (and website, now) called "cdecl" to help with these, it says:

• (char *(*)()): cast unknown_name into pointer to function returning pointer to char
• (int (*)()): cast unknown_name into pointer to function returning int

Answer 2

The easiest way of deciphering complex C expressions is to start with the innermost expression, then in an anti-clockwise pattern move on to the next. (int (*)())

1. (*) A Pointer, anti-clockwise motion hits on (, then move on again to hit on )
2. (*)() A pointer to function, move on to (, then move on again to )
3. int (*)() A pointer to a function returning int, move on to int, then move on to hit on )
4. (int (*)()) finally move on to (, and there you have it,

A pointer to a function returning int, since it is wrapped in the outer () is because of the macro.

Hope this helps, Best regards, Tom