Call a void* as a function without declaring a function pointer

Question

I\'ve searched but couldn\'t find any results (my terminology may be off) so forgive me if this has been asked before.

I was wondering if there is an easy way to call a

Answer 1

I get awfully confused when casting to function types. It's easier and more readable to typedef the function pointer type:

void *ptr = ...;
typedef void (*void_f)(void);
((void_f)ptr)();

Answer 2

In C++: reinterpret_cast< void(*)() > (ptr) ()

The use of reinterpret_cast saves you a set of confusing parentheses, and the < > clearly sets the type apart from the call itself.

Answer 3

Your cast should be:

((void (*)(void)) ptr)();

In general, this can be made simpler by creating a typedef for the function pointer type:

typedef void (*func_type)(void);
((func_type) ptr)();

I should, however, point out that casting an ordinary pointer (pointer to object) to or from a function pointer is not strictly legal in standard C (although it is a common extension).