Question

I’ve got these prototypes:

int *my_func(int x, void (*other_func)(int a, int b));

int func2(int val1, int val2);

Assume there is a function written that matches it.

If I want to actually call my_func, how would I do it? I’ve tried the following with no luck:

my_func(1,func2);

Here’s the error:

warning: passing argument 2 of ‘my_func’ from incompatible pointer type

Answer 1

That’s because the function prototypes don’t match:

void (*other_func)(void *a, int *b)

is not the same as:

int func2(int val1, int val2);

One takes a void* and int*, while the other takes two ints.

EDIT:

Also, the return type doesn’t match.

EDIT 2:

Since you’ve fixed both errors, this answer is out of context. I just tested it with these two fixes, and it compiles:

int *my_func(int x, void (*other_func)(int a, int b)){
    return 0;
}
void func2(int val1, int val2){
    printf("blah");
}


int main(){
    my_func(1,func2);

    return 0;
}