Re: Multiple functions applied to a single value

[Graham Klyne]

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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