Re: if expression and match expression

Sébastien and Gabriel;

Thank you for your reply.

Got it. My original goal was creating has_element function without if
expression.

And seems it is the one I was looking for;

let rec has_element2 l e =
match l with
| [] -> false
| h::t when h = e -> true
| _ -> has_element2 (List.tl l) e      
;;


Thank you for all your help!
Thanks to Hendrik as well :-) 

Regards,

Seungjin 


On 10/09 09:49 AM, Sébastien Dailly [email protected] [ocaml_beginners] wrote:
> 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

Seungjin Kim