Hello.

I think you are confusing the operators = and ==. = is used to check if two objects have the same "logical" contents whether == checks if two objects are exactly the same in memory.
When you type a = insert a "word", the function insert returns a new lex_tree which has the same contents as the variable a but is another object in memory. That is why I think you wanted to write :

let a = insert [] "word"
in
a = insert a "word" ;;

which effectively returns true. It is important to understand that the variable a and the result returned by the function insert are logically the same but not physically.

Samuel.

On Sun, 23 Feb 2003 17:29:14 +0100
"Alexander S. Usov" <A.S.Usov@KVI.nl> wrote:

> Hi!
> 
> Can anybody expalin me why in such code
> 
> ------
> type lex_node = Letter of char * bool * lex_tree
> and  lex_tree = lex_node list ;;
> 
> exceprion Already ;;
> 
> let rec insert lex word =
>   let whd = word.[0]
>   and wtl = String.sub word 1 (String.length word - 1)
>   in
>     try
>       match lex with
> 	| (Letter(c,b,t) as letter)::tail when (c <> whd) -> 
> 	    letter :: (insert tail word)
> 	| Letter(c,b,t) :: tail when (wtl = "") ->
> 	    if b = true then
> 	      raise Already
> 	    else
> 	      Letter(c,true,t) :: tail
> 	| Letter(c,b,t) :: tail ->
> 	    Letter(c,b,(insert t wtl)) :: tail
> 	| [] ->
> 	    if wtl = "" then
> 	      [ Letter (whd, true, []) ]
> 	    else
> 	      [ Letter (whd, false, insert [] wtl) ]
>     with
>       | Already -> lex ;;
> 
> let a = insert [] "word"
> in
>   a == insert a "word" ;;
> -----
> 
> returns false?
> 
> -- 
> Best regards,
>   Alexander.
> 
> 
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-- 
Samuel Mimram

Samuel.Mimram@ens-lyon.fr
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