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

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