On Friday 03 August 2007 00:17:50 Msam85 wrote:
> HI everyone.
>
> I'm new in Ocaml language, and I had tried to understand the following
> function, but Im not able.
>
> let op = function
> p->fst p (snd p);;
This is equivalent to:
let op p = fst p (snd p)
Pattern matching makes deconstruction faster and more readable, so it is even
easier to write:
let op (f, x) = f x
So the "op" function accepts a pair and applies the second element of the pair
to the first element of the pair.
> If I put it at the top level it shows me :
> val op : ('a->'b)*'a ->'b = <fun>
>
> I think that p must be like ( , ) but if i trie to write op((+),3);;
> It shows me -: int -> int = <fun>
Yes. That takes the curried "add" function and partially applies "3" to give a
function that adds three:
# let op (f, x) = f x;;
val op : ('a -> 'b) * 'a -> 'b = <fun>
# let add_three = op((+), 3);;
val add_three : int -> int = <fun>
# add_three 5;;
- : int = 8
--
Dr Jon D Harrop, Flying Frog Consultancy Ltd.
OCaml for Scientists
http://www.ffconsultancy.com/products/ocaml_for_scientists/?e
