* Re: Streams in Caml Light 0.5
1992-10-26 0:33 Streams in Caml Light 0.5 HEDDEN
@ 1992-10-26 16:10 ` Michel Mauny
0 siblings, 0 replies; 2+ messages in thread
From: Michel Mauny @ 1992-10-26 16:10 UTC (permalink / raw)
To: HEDDEN; +Cc: caml-list
> 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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