OCaml function type ('a - 'b) - 'a - 'b

Question

I'm having some problems understanding how a function of this type works. ``` let try_it f x = f x;; val try_it : ('a -> 'b) -> 'a -> 'b = ``` The above function, try_it accepts a function f which accepts a polymorphic parameter and returns a polymorphic type. Now I can easily pass functions + parameters to try_it of varying number of parameters like below and it works. ``` let add_three a b c = a + b + c;; let add_two a b = a + b;; let try_it f x = f x;; print_endline(string_of_int(try_it add_two 23 45));; print_endline(string_of_int(try_it add_three 23 4 56));; ``` Now if the above works, why can't I create a function called try_it2 which is defined like below and pass it the same functions and parameters but in reverse order? ``` let try_it2 x f = f x;; val tryit2 : 'a -> ('a -> 'b) -> 'b = ``` To me try_it2 is the same as try_it with only the parameters order changed but I'm obviously missing something here. Can someone please put me on the correct path?

Accepted Answer

The real problem is that the number of arguments of your try_it function is fixed.

You must understand correctly what goes on on this line:

```
print_endline(string_of_int(try_it add_two 23 45));;
```

try_it add_two 23 returns a function of type int -> int, which is in turn evaluated over the value 45, giving 23 + 45.

but if you change the definition of try_it as you did, then this line becomes:

```
print_endline(string_of_int(try_it 23 add_two 45));;
```

So that try_it 23 add_two returns a (int -> int) function, evaluated over 45.

Have a nice day !

OCaml function type ('a -> 'b) -> 'a -> 'b

Solution

Problem