From here: ISO C Void * and Function Pointers, I have found a workaround to cast (void*)
to function pointer:
int
main(int argc, char *argv[])
{
...
void (*funcp)(void); /* Pointer to function with no arguments */
...
*(void **) (&funcp) = dlsym(libHandle, argv[2]);
}
In other words - to dereference double pointer (another level of indirection).
Now that still assumes, that the final function is of void(*)()
type, but I would like to make cast available to other function "types" that can for example accepts some arguments.
Then I found another workaround how to wrap function pointer in struct Why can't I cast a function pointer to (void *)? :
typedef struct
{
void (*ptr)(void);
} Func;
Func vf = { voidfunc };
So I would like to merge these 2 ideas and make possible to pass arbitrary function type as function pointer via struct:
#include <stdio.h>
struct s{
int a, b;
void (*func)();
};
typedef struct{
int (*add)(int,int);
} Func;
int add(int a, int b) { return a+b; }
int main(){
Func f = {add};
struct s foo = {.func=(void*)(&f)};
printf("%i\n",f.add(1,2));
printf("%i\n",foo.func(1,2));
}
Unfortunately, it gives the error:
invalid use of void expression
So the question is, how to cast back the type (void*)
to (int*)(int,int)
inside of the printf
statement?
even if you change the function to return int
(int (*func)();) and eventually it will compile, your code is wrong.
Calling the function pointer is in the fact a dereferencing of this pointer.
When you assign the function pointer with the address of the struct, calling this function will actually execute the data inside the struct - not the function referenced struct member. It of course will not be successful.
The following example is an UB but works on x86 & arm machines and it is only for the illustrational purposes..
struct s{
int a, b;
int (**func)();
};
typedef struct{
int (*add)(int,int);
} Func;
int add(int a, int b) { return a+b; }
int main(){
Func f = {add};
struct s foo = {.func=(void*)(&f)};
printf("%i\n",f.add(1,2));
printf("%i\n",(*foo.func)(1,2));
}
or if you want to use the void (**)()
pointer in struct
typedef int func();
struct s{
int a, b;
void (**func)();
};
typedef struct{
int (*add)(int,int);
} Func;
int add(int a, int b) { return a+b; }
int main(){
Func f = {add};
struct s foo = {.func=(void*)(&f)};
printf("%i\n",f.add(1,2));
printf("%i\n",((func *)(*foo.func))(1,2));
}
or
typedef int func();
struct s{
int a, b;
void (*func)();
};
typedef struct{
int (*add)(int,int);
} Func;
int add(int a, int b) { return a+b; }
int main(){
Func f = {add};
struct s foo = {.func=(void*)(&f)};
printf("%i\n",f.add(1,2));
printf("%i\n",(*((func **)foo.func))(1,2));
}
or without typedefs
struct s{
int a, b;
void (*func)();
};
typedef struct{
int (*add)(int,int);
} Func;
int add(int a, int b) { return a+b; }
int main(){
Func f = {add};
struct s foo = {.func=(void*)(&f)};
printf("%i\n",f.add(1,2));
printf("%i\n",(*((int (**)())foo.func))(1,2));
}