Daniel Schuessler wrote:
> The thing I don't understand yet is the last line: Why is it OK to discard the
> leftover input from the (f x) Iteratee and yield just the leftover input from
> the first one (m0)?
First of all, the question is about an older version of Iteratee. For
example, the following code
http://okmij.org/ftp/Haskell/Iteratee/Iteratee.hs
defines Iteratee a bit differently so the question does not apply.
> data Iteratee a = IE_done a
> | IE_cont (Maybe ErrMsg) (Stream -> (Iteratee a,Stream))
>
> instance Monad Iteratee where
> return = IE_done
> IE_done a >>= f = f a
> IE_cont e k >>= f = IE_cont e (docase . k)
> where
> docase (IE_done a, stream) = case f a of
> IE_cont Nothing k -> k stream
> i -> (i,stream)
> docase (i, s) = (i >>= f, s)
No left-over is discarded any more.
Your question is about the previous design, called `The second design'
described in Iteratee.hs. The corresponding comment block answers your
question, please search for ``Justification for the case IE_done x
s >>= f''.
Please see
http://okmij.org/ftp/Haskell/Iteratee/IterateeM.hs
for the `production' case of Iteratee in a base monad. The file
IterateeM.hs talks about one more design, and its drawbacks.
It's all about finding the optimal trade-off, I guess.
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