On 8/3/07, Neil Mitchell <[EMAIL PROTECTED]> wrote:
> 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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