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Re: Streams in Caml Light 0.5
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Date: -- (:)
From: Michel.Mauny@i...
Subject: Re: Streams in Caml Light 0.5
> I am having a small difficulty using streams in Caml Light v0.5.
> 
> The example below might be a bug, but is more likely due to a
> lack of understanding on my part.
> 
> The exception in the following toplevel session indicates that
> there is a problem:
> 
> 
>         >       Caml Light version 0.5
> 
>         #include "test";;
>         map_stream : ('a -> 'b) -> 'a stream -> 'b stream = <fun>
>         filter_stream : ('a -> bool) -> 'a stream -> 'a stream = <fun>
>         int_stream : int stream = <abstract>
>         - : int = 1
>         - : int = 9
>         - : int = 25
>         odd_sq_stream : int stream = <abstract>
>         - : int = 49
>         Uncaught exception: Parse_failure
>         - : unit = ()
>         #quit();;

The problem is that your map_stream and filter_stream are incorrect
(if I understand correctly what they should do) they both return a
stream with only one element.  Your int_stream (correctly) contains
the stream of strictly positive integers.  Considering that a
successive match on a stream *physically* destroys the stream. Now,
what you do is select the first odd element (1), compute its square,
and return a stream with that single element.  The next time you do
it, it returns 9 and so on. These calls succeed because filter_stream
builds a new stream (with only one element) each time Now, if you
encapsulate the process into a function, map_stream empties the result
of filter_stream after the first call, and fails (the stream is empty)
for subsequent calls. Cf. the comments and corrections below.

> The file "test.ml" (that is loaded above) contains the following:
> 
> 
>         (* Maps a function over a stream *)
>         let rec map_stream func = function
>             [< 'x >]  ->  [< 'func x >]
>           | [< >]     ->  [< >]
>         ;;

I guess your map_stream function is supposed to build the stream of
results of applying a function `func' on *all* elements of the
argument stream. In that case, the correct definition should be:

        let rec map_stream func = function
                  [< 'x ; (map_stream func) str >] -> [< '(func x); str >]
                | [< >] -> [< >]
        ;;

You missed the recursive call.

>         (* Filters a stream according to a predicate *)
>         let rec filter_stream pred strm =
>           match strm with
>               [< 'x >]  ->  if pred x then [< 'x >]
>                                       else filter_stream pred strm
>             | [< >]     ->  [< >]
>         ;;

Now, filter_stream is also wrong, for the same reason (a resursive
call is missing). The correct definition should be:

        let rec filter_stream pred = function
                  [< 'x ; (filter_stream pred) str >]
                        -> if pred x then [< 'x ; str >] else str
                | [< >] -> [< >]
        ;;

For streams, it is very common that recursive calls occur in the
`stream pattern' (this is by analogy with BNF grammars, cf. the part
of the documentation dedicated to streams and parsers).

>         (* The stream of positive integers *)
>         let int_stream = ints 1
>           where rec ints n = [< 'n; ints (succ n) >]
>         ;;
> 
>         (* Use `filter_stream' and `map_stream' to create a
>            stream of the squares of the odd integers *)
>         (* These work correctly *)
>         stream_next (map_stream (function x->x*x)
>                      (filter_stream (function n->(n mod 2)= 1)
>                      int_stream));;
>         stream_next (map_stream (function x->x*x)
>                      (filter_stream (function n->(n mod 2)= 1)
>                      int_stream));;
>         stream_next (map_stream (function x->x*x)
>                      (filter_stream (function n->(n mod 2)= 1)
>                      int_stream));;

Above, all calls succeeded because filter_stream builds a new
`singleton' stream, map_stream returns another singleton stream, and
you requested only for its first element.

>         (* Encapsulate the above *)
>         let odd_sq_stream =
>                 map_stream (function x->x*x)
>                 (filter_stream (function n->(n mod 2)= 1)
>                 int_stream)
>         ;;
> 
>         (* Test the above encapsulation *)
>         (* This works correctly *)
>         stream_next odd_sq_stream;;
> 
>         (* This causes an error *)
>         stream_next odd_sq_stream;;

The encapsulation `odd_sq_stream' of the stream contains only one
element, this is why the second call failed.

Best regards,

Michel
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