Why do i need to remove the function pointer casting to use the function as shown below ?
This Compiles:
#include <stdio.h>
int print_int(int i){
printf("%d", i);
return i;
}
typedef int (*print_int_func) (int);
int main(){
void** p = malloc(1*sizeof(print_int_func*));
p[0] = (print_int_func*)print_int;
((print_int_func)p[0])(2); // This Compiles
((print_int_func*)p[0])(2); // This does NOT
return 0;
}
The declaration typedef int (*print_int_func) (int); declares print_int_func to be a pointer to a specific type of function. So (print_int_func)p[0] casts p[0] to such a pointer to a function, but (print_int_func*)p[0] casts p[0] to a pointer to a pointer to a function. Thus, the result is a pointer to an object (that object being a pointer to a function). Since it is an object, not a function (or pointer to a function), it cannot be called like a function.
Additionally, avoid using void * for pointers to functions. void * is a pointer to an object, and the C standard does not define conversions between pointers to objects and pointers to functions. To create a “generic” pointer to a function, simply choose any function type and use a pointer to that type:
For example, you can declare an arbitrary type:
typedef void (*CommonFunctionPointer)(void);
and make an array of them:
CommonFunctionPointer *p = malloc(N * sizeof *p);,
and then you can store any function pointer in the array:
p[i] = (CommonFunctionPointer) print_int;
and use a pointer from the array by casting it back to its correct type:
((int (*)(int)) p[i])(2);.
(print_int_func)p[0] casts p[0] into a function so it can be called.
p[0] to a pointer to a function. The function call operator (parentheses after an expression) only calls pointers to functions. When you write a function call using a function name, like sin(x), the function is automatically converted to a pointer.q that points to a function and write (*q)(x), *q is a function, but it is automatically converted to a pointer to the function. And then you can apply * again. (**q)(x) calls the same function. So does (******q)(x).)
void**, you need to usemalloc(1 * sizeof(void*)). The argument tosizeofis always the destination type with one less*.void*can hold a function pointer.void *in the first place; function pointers are different from object pointers, andvoid *is not a generic type for function pointers. Additionally, its preferable to use the actual identifier being assigned, with an*, for thesizeof, rather than repeating the type:AnyType *p = malloc(NumberOf * sizeof *p);.