[Haskell-cafe] Re: monad subexpressions

[david48]

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

This is how I understand it:

> Can you use (<-) outside of a do block?
> b >> f (<- a)

b >> do { ta <-a; f ta }
or
b >> a >>= \ta -> f ta

> What are the semantics of
> do b >> f (<- a)

do b >> a >>= \ta -> f ta

> Given:
>
> if (<- a) then f (<- b) else g (<- c)

a >>= \ta -> if (ta) then ( b >>= \tb -> f tb ) else ( c >>= \tc -> f tc )

> do let x = (<- a)
>     f x

No idea if that could be possible. or maybe :

do a >>= \ta -> let x = ta in f x


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