[Haskell-cafe] trying to understand space leaks....

[Daryoush Mehrtash]

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 at gmail.com> wrote:

> On Tue, May 26, 2009 at 5:03 PM, Daryoush Mehrtash <dmehrtash at 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
>
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