Multiple functions applied to a single value
Derek Elkins
ddarius at hotpop.com
Tue Dec 9 08:00:45 EST 2003
On Mon, 08 Dec 2003 16:33:53 +0000
Graham Klyne <GK at ninebynine.org> wrote:
> A little while ago, I sent a long rambling message [1] containing a
> key question in response to an earlier helpful response. I guess my
> ramblings were quite reasonably passed over as too much waffle, and
> the question was lost. Here, I isolate the question.
>
> From the earlier response, I understand that:
>
> ((->) r)
>
> is an instance of Monad, but I haven't found a definition of the Monad
> instance functions. I think these do the job, and they satisfy the
> monad laws [2]:
>
> [[
> instance Monad ((->) r) where
> return a = const a
> g1 >>= f = \e -> f (g1 e) e
> ]]
>
> Can anyone confirm, or point me at an actual definition?
http://www.haskell.org/hawiki/MonadReader
There's also the source of Control.Monad.Reader (and the new library) as
well as various papers containing implementations.
And, as long as you didn't make a mistake below, you've proven it.
Well, you've proven it's a monad in this way.
> [2] Checking the Monad laws for the above definitions
> (cf. Haskell report p90):
>
> (A)
> return a >>= k
> = [g1/const a][f/k] \e -> f (g1 e) e
> = \e -> k (const a e) e
> = \e -> k a e
> = k a -- as required.
>
> (B)
> m >>= return
> = [g1/m][f/const] \e -> f (g1 e) e
> = \e -> const (m e) e
> = \e -> (m e)
> = m -- as required.
>
> (C)
> m >>= (\x -> k x >>= h)
> = [g1/m][f/(\x -> k x >>= h)] \e -> f (g1 e) e
> = \e -> (\x -> k x >>= h) (m e) e
> = \e -> (k (m e) >>= h) e
> = \e -> ([g1/k (m e)][f/h] \e1 -> f (g1 e1) e1) e
> = \e -> (\e1 -> h (k (m e) e1) e1) e
> = \e -> h (k (m e) e) e
>
> (m >>= k) >>= h
> = ([g1/m][f/k] \e -> f (g1 e) e) >>= h
> = (\e -> k (m e) e) >>= h
> = [g1/(\e -> k (m e) e)][f/h] \e1 -> f (g1 e1) e1
> = \e1 -> h ((\e -> k (m e) e) e1) e1
> = \e1 -> h (k (m e1) e1) e1
> = \e -> h (k (m e) e) e
>
> So both expressions are equivalent, as required.
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