Multiple functions applied to a single value
Graham Klyne
GK at ninebynine.org
Mon Dec 8 16:33:53 EST 2003
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?
#g
--
[1] http://haskell.org/pipermail/haskell-cafe/2003-November/005566.html
[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.
------------
Graham Klyne
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