I (think)  I understand the problem.  What I don't have any intuition about is how much  space would "Expensive Structure" take if it was basically an IO Char computation fed into a simple function (say checks for char being equal to "a").   Is there any way to guess, know the size of the buffer that is kept in the heap?

thanks,

Daryoush
On Wed, May 27, 2009 at 3:12 PM, Ryan Ingram <ryani.spam@gmail.com> wrote:
There's still the space used by the closure "b".

An example:

expensiveParser :: Parser Char ExpensiveStructure

simple :: Parser Char Int

withExpensive :: ExpensiveStructure -> Parser Char Int
withExpensive _ = mzero  -- actually always fails, not using its argument.

example = do
   e <- expensiveParser
   simple `mplus` withExpensive e

The expensive structure constructed by expensiveParser needs to be
kept in memory throughout the entire parsing of "simple", even though
withExpensive doesn't actually use it and would immediately fail.  A
smarter parser could realize that e couldn't actually ever be used and
allow the GC to free it much more quickly.

This example can be made arbitrarily more complicated; withExpensive
could run different things based on the value of "e" that could be
determined to fail quickly, simple might actually do a lot of work,
etc.  But during the "mplus" in the monadic parser, we can't free e.

 -- ryan

On Wed, May 27, 2009 at 12:49 PM, Daryoush Mehrtash <dmehrtash@gmail.com> wrote:
> So long as the [s] is a fixed list (say [1,2,3,4]) there is no space
> leak.    My understanding was that the space leak only happens if there is
> computation involved in building the list of s.      Am I correct?
>
> If so, I still don't have any feeling for what needs to be saved on the heap
> to be able to back track on computation that needs and  IO computation
> data.    What would be approximate  space that an IO (Char) computation
> take  on the heap, is it few bytes, 100, 1k,  ....?
>
> Daryoush
>
>
> On Tue, May 26, 2009 at 6:11 PM, Ryan Ingram <ryani.spam@gmail.com> wrote:
>>
>> On Tue, May 26, 2009 at 5:03 PM, Daryoush Mehrtash <dmehrtash@gmail.com>
>> wrote:
>> > newtype Parser s a = P( [s] -> Maybe (a, [s]))
>> (fixed typo)
>>
>> > instance MonadPlus  Parser where
>> >       P a mplus P b = P (\s -> case a s of
>> >                             Just (x, s') -> Just (x, s')
>> >                             Nothing -> b s)
>>
>> > a)what exactly gets saved on the heap between the mplus calls?
>>
>> Two things:
>>
>> (1) Values in the input stream that "a" parses before failing.
>> Beforehand, it might just be a thunk that generates the list lazily in
>> some fashion.
>>
>> (2) The state of the closure "b"; if parser "a" fails, we need to be
>> able to run "b"; that could use an arbitrary amount of space depending
>> on what data it keeps alive.
>>
>> > b)I am assuming the computation to get the next character for parsing to
>> > be
>> > an "IO Char" type computation,  in that case, what would be the size of
>> > the
>> > heap buffer that is kept around in case the computation result needs to
>> > be
>> > reused?
>>
>> Nope, no IO involved; just look at the types:
>>
>> P :: ([s] -> Maybe (a,[s])) -> Parser s a
>>
>> (Parser s a) is just a function that takes a list of "s", and possibly
>> returns a value of type "a" and another list [s] (of the remaining
>> tokens, one hopes)
>>
>> It's up to the caller of the parsing function to provide the token
>> stream [s] somehow.
>>
>> > c) Assuming Pa in the above code reads n tokens from the input stream
>> > then
>> > fails, how does the run time returns the same token to the P b?
>>
>> It just passes the same stream to both.  No mutability means no danger :)
>>
>>  -- ryan
>
>
>