Re: if expression and match expression

Le 2015-10-09 05:55, Seung-jin Kim [email protected] 
[ocaml_beginners] a écrit :
> First of all,
> There was a typo in my very initial question.
> my has_element1 should be
> 
> let rec has_element1 l e =
> match l with
> | [] -> false
> | h::t -> if e = h then true else has_element1 t e;;
> 
> Anyway,, Seems everyone got my point. :-) First time to post this
> group and very new to ocaml.
> 
> I did with ( ) for my second match.
> 
> utop[91]> let rec has_element2 l e =
> match l with
> | [] -> false
> | h::t -> ( match h with
> | e -> true
> | _ -> has_element2 t e
> )
> ;;
> 
> val has_element2 : 'a list -> 'b -> bool = <fun>
> Characters 107-108:
> Warning 11: this match case is unused.
> utop[92]> has_element2 [2;3;4] 10;;
> - : bool = true
> utop[93]>
> 
> Still getting the same warning message with the same result.

Hello,

when you write

> match l with
> | [] -> false
> | h::t -> …

You do not match l with an existing variables named h and t. You create 
two new ones which match the pattern.

The same applies when you write :

> match h with
> | e -> true

You create a new variable name « e » which override the existing 
variable. Of course this pattern always match.

You can write :

- either an if / else structure as in your first example.
- either a gard pattern in your pattern matching.

Regards