On Thu, Feb 12, 2009 at 6:36 PM, Edsko de Vries
<devriese@cs.tcd.ie> wrote:
Hi,
I can desugar
do x' <- x
f x'
as
x >>= \x -> f x'
which is clearly the same as
x >>= f
However, now consider
do x' <- x
y' <- y
f x' y'
desugared, this is
x >>= \x -> y >>= \y' -> f x' y'
I can simplify the second half to
x >>= \x -> y >>= f x'
but now we are stuck. I feel it should be possible to write something like
x ... y ... f
or perhaps
f ... x ... y
the best I could come up with was
join $ return f `ap` x `ap` y
which is not terrible but quite as easy as I feel this should be. Any hints?
Edsko
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