Question

If I have a prototype that is declared as:

void foo(int (*fi)(void *, void *))

And I call the function with something like this:

foo(int (*)(void*, void*)(bool_expr ? func1 : func2));

Where func1 and func2 are defined as follows:

int func1(char *s1, char *s2);
int func2(int *i1, int *i2);

Is the function call casting one of the functions (func1 ^ func2) to the type of the function required by foo? Could I also make the prototype for foo look like:

void foo(int (*)(void *, void *))

Answer 1

As commented the code as posted did not compile.

FWIW, this compiles and executes as you would expect with VC2010:

int func1(char *s1, char *s2) { printf("func1\n"); return 0; }
int func2(int *i1, int *i2)   { printf("func2\n"); return 0; }

void foo(int (*)(void *, void *));

int main(int argc, char** argv)
{
    foo(2 == argc ? (int(*)(void*,void*))func1 :
                    (int(*)(void*,void*))func2);

    return 0;
}

void foo(int (*a)(void *, void *))
{
    a((void*)0, (void*)0);
}