Hello Mauri­cio, Friday, August 29, 2008, 5:41:41 PM, you wrote: afaik, this shorthand isn't exact. actually there is more code dealing with fails and this code use Monad operations

Hi,

http://haskell.org/haskellwiki/Keywords says that:

------------- [do is a] syntactic sugar for use with monadic expressions. For example:

do { x ; result <- y ; foo result }

is shorthand for:

x >> y >>= \result -> foo result -------------

I did some tests hiding Prelude.>> and Prelude.>>= and applying >>> and >>= to non-monadic types, and saw that 'do' would not apply to them. So, I would like to add the following to that text:

------------- as long as proper types apply:

x :: Prelude.Monad a y :: Prelude.Monad b foo :: b ->> Prelude.Monad c -------------

Is that correct (Haskell and English)?

Thanks, Mauricio

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On Fri, Aug 29, 2008 at 6:41 AM, Maurí­cio wrote:

Hi,

http://haskell.org/haskellwiki/Keywords says that:

------------- [do is a] syntactic sugar for use with monadic expressions. For example:

do { x ; result <- y ; foo result }

is shorthand for:

x >> y >>= \result -> foo result -------------

I did some tests hiding Prelude.>> and Prelude.>>= and applying >> and >>= to non-monadic types, and saw that 'do' would not apply to them. So, I would like to add the following to that text:

It sounds like you tried to redefine (>>) and (>>=) and make 'do' use the new definitions. This is not possible, regardless of what types you give (>>) and (>>=). If you want to define (>>) and (>>=), do so for a particular instance of Monad.

------------- as long as proper types apply:

x :: Prelude.Monad a y :: Prelude.Monad b foo :: b -> Prelude.Monad c -------------

Is that correct (Haskell and English)?

Thanks, Maurício

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