3 Aug
2007
3 Aug
'07
4:53 p.m.
On 8/3/07, Neil Mitchell
temp <- a let x = temp
if you write : let x = (<-a):x is it possible that is desugars into : temp <-a let x = temp:x that would'nt work ? I realize I may be asking dumb questions but being dumb never harmed anyone so :) Also :
do case x of [] -> return 1 (y:ys) -> f (<- g y)
Is it not possible that is desugars to do case x of [] -> return 1 (y:ys) -> g y >>= \temp -> f temp
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