Re: [Haskell-beginners] Monad and bind operator interpretation

Yeah, after my first couple of significant Haskell projects I came to the conclusion that having (>>=) instead of (=<<) as the "canonical bind" is a wart. (>>=) makes for clean desugaring of do notation, but obscures that (a -> m b) -> (m a -> m b) intuition, which I think is more important pedagogically and also tends to be cleaner in non-do monadic code. To a beginner, flip is an easy mechanical concept, but analyzing the resultant type for new insights is not a habit yet. The one whose purpose is purely mechanical should be "the flipped one." On Dec 14, 2015 10:41 AM, "Joel Williamson" wrote:

This function is actually in the Prelude as (=<<).

On Mon, 14 Dec 2015, 13:17 Dániel Arató wrote:

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On 14/12/2015, Raja wrote:

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So extending this interpretation - can I swap the two parameters (?)

Now my new hypothetical interpretation becomes:

(>>=) :: (a -> m b) -> m a -> m b

Sure, bind' :: Monad m => (a -> m b) -> m a -> m b bind' = flip (>>=)

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If i further add parens:

(>>=) :: (a -> m b) -> (m a -> m b)

Yeah, that's exactly the same thing. Types are right associative. _______________________________________________ Beginners mailing list Beginners@haskell.org http://mail.haskell.org/cgi-bin/mailman/listinfo/beginners

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