Reputation: 650
I can't understand why the following function works with 2 arguments even if we declare it with one param:
let rec removeFromList e = function
h :: t -> if h=e then h
else h :: removeFromList e t
| _ -> [];;
removeFromList 1 [1;2;3];;
Upvotes: 3
Views: 411
Reputation: 35210
You're declaring it with two parameters. The syntax:
let f = function ...
can be seen as a shortcut for
let f x = match x with
So, your definition is actually:
let rec removeFromList e lst = match lst with
h :: t -> if h=e then h else h :: removeFromList e
Upvotes: 4