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Today's Topics:
1. Re: f . g or f g or f $ g? (Ertugrul S?ylemez)
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Message: 1
Date: Wed, 13 Feb 2013 09:12:11 +0100
From: Ertugrul S?ylemez <[email protected]>
Subject: Re: [Haskell-beginners] f . g or f g or f $ g?
To: [email protected]
Message-ID: <[email protected]>
Content-Type: text/plain; charset="us-ascii"
Mateusz Kowalczyk <[email protected]> wrote:
> A less obvious interpretation is to treat ($) as `id'.
>
> (f . g) x = f $ g x = f (id g x) = f (g x)
This is not how you get from ($) to id. The correct path is:
f $ x = ($) f x = f x = id f x = f `id` x
This equivalence is indicated by the type of ($). It's a specialized
instance of a -> a:
($) :: (a -> b) -> (a -> b)
($) f = f
or equivalently:
($) :: (a -> b) -> a -> b
($) f x = f x
or equivalently:
($) :: (a ~ b -> c) => a -> a
($) = id
Greets,
Ertugrul
--
Not to be or to be and (not to be or to be and (not to be or to be and
(not to be or to be and ... that is the list monad.
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