Re: if expression and match expression

On Thu, Oct 08, 2015 at 08:12:48PM -0700, [email protected] [ocaml_beginners] wrote:
> Code first:
>  
> 
>  let rec has_element1 l e =
> match l with
> | [] -> false
> | h::t -> if e = h then true else false
> ;;
> 
> let rec has_element2 l e =
> match l with
> | [] -> false
> | h::t -> match h with
>           | e -> true
>       | _ -> has_element2 (List.tl l) e
> ;;
> 
>  
> 
>  has_element1 works as I expected. 
> 
>  And I rewrite has_element1 with replacing 'if expressing' with 'match. 
> 
>  But has_element2 always return 'true' and it gives me warning: 'Warning 11: this match case is unused'
>  
> 
>  Questions: 
> 
>   (1) Why does my has_element2 always return true? Match expression I 
> used in has_element2 not the same as if expression in has_element1?

How is the poor compiler to know whether the last alternative belongs 
with the first or the second match?  Try inserting ( before the second 
match keyword and putting its ")" where it belongs.

I think this may be a language design problem.  But the problem ends up 
being caught in the type analysis, so it's probably only confusing.

When OCaml misbehaves I usually find it enlightening to be explicit 
about types and parentheses.

-- hendrik