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

28 May 2009


      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  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 
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 
wrote:
...
On Tue, May 26, 2009 at 5:03 PM, Daryoush Mehrtash 
...
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