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

c function pointers void
jay.lee picture jay.lee · Apr 14, 2010 · Viewed 14.8k times · Source

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 void* as a function in C without first declaring a function pointer and then assigning the function pointer the address;

ie. assuming the function to be called is type void(void)

void *ptr;
ptr = <some address>;
((void*())ptr)(); /* call ptr as function here */

with the above code, I get error C2066: cast to function type is illegal in VC2008

If this is possible, how would the syntax differ for functions with return types and multiple parameters?

Answer

jamesdlin picture jamesdlin · Apr 14, 2010

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).

Related questions

How do you pass a function as a parameter in C?

I want to create a function that performs a function passed by parameter on a set of data. How do you pass a function as a parameter in C?

Read More
C Programming: malloc() inside another function

I need help with malloc() inside another function. I'm passing a pointer and size to the function from my main() and I would like to allocate memory for that pointer dynamically using malloc() from inside that called function, but what …

Read More
Function pointers and address of a function

So I figured when making function pointers, you do not need the operator & to get the address of the initial function: #include <stdio.h> double foo (double x){ return x*x; } int main () { double (*fun1)(double) = &…

Read More