[Haskell-cafe] Re: Re: Re: monad subexpressions

[david48]

On 8/3/07, Neil Mitchell <ndmitchell at gmail.com> wrote:

> > Is it not possible that is desugars to

> > do case x of
> >          [] -> return 1
> >          (y:ys) -> g y >>= \temp -> f temp

> See the rule about always binding to the previous line of a do block.
> This case then violates that.

I assumed that the example was equivalent to :

do case x of
         [] -> return 1
         (y:ys) -> do f (<- g y)

Shouldn't the rule work then ?