Function pointer expression

Question

So I've got the following expression:

int (*f1(int(*a)(int, int))) (int, int);

and I'm trying to make sense out of it, but it's confusing. I figured out that "a" is a pointer to a function which takes 2 arguments(int, int). Then f1 seems to be a pointer to another function that takes 2 int arguments. But what's confusing me is how f1 relates to a.

Can someone give me some hints or properly explain what the above expression is doing?

Answer 1

The tip in C is to read a declaration as it were an expression. This is this famous symmetrie that make C elegant.

How to read it? following the operators precedence rules:

1. *a: if I dereference variable a;
2. (*a)(int,int): and then call it with two integers;
3. int (*a)(int,int): then I get an integer;

So a is a pointer to a function taking two ints as parameter and returning an int.

Then:

5. f( int(*a)(int,int) ) if I call f with the argument a;
6. *f( int(*a)(int,int) ) and then I dereference the result;
7. (*f( int(*a)(int,int) )(int,int) and then call this result with 2 int as argument
8. int (*f( int(*a)(int,int) )(int,int) I get an int

So f is a function taking an a as argument and returning a pointer to a function that take two ints as argument and returning an int. So f return type is the same as its argument return type. It could have been simpler:

using ftype = int(*)(int,int);
ftype f( ftype a);