Passing function as argument does not work

Question

I want to pass a function as argument to another function. I searched already Google about information on this and I found already a explanation but it doesn't work for me and I don't know why.

I have following code:

void doSomething(uint64_t *);

This is my function I want to pass.

int functionToCall(int x, int y, void (*f)(uint64_t *));

This is my function I want to call and pass the doSomething() function.

In my code now is:

uint64_t *state = malloc(sizeof(uint64_t) * 10);
void (*f)(uint64_t *) = doSomething;
functionToCall(2, 3, f(state));

If I compile the above code now I get always:

error: invalid use of void expression

What is the problem with that?

Answer 1

The error come from the fact that you dont pass a pointer to the function but the function's result (which is void).

If in your function functionToCall you want to call doSomething with the state variable, then you should do something like this:

void doSomething(uint64_t *);
int functionToCall(int x, int y, unint64_t * state, void (*f)(uint64_t *))
{
    f(state);

    /* ... */
}

uint64_t *state = malloc(sizeof(uint64_t) * 10);
void (*f)(uint64_t *) = doSomething;
functionToCall(2, 3, state, f);    

Answer 2

I did not realize that I have passed the argument for the doSomething() function in my functionToCall function.

So, the line,

functionToCall(2, 3, f(state));

Has to be,

functionToCall(2, 3, f);

Answer 3

I'm not that good in C but i guess that you could get some help from previous similar post

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

Hope that my answer helped