On Mon, 08 Dec 2003 16:33:53 +0000 Graham Klyne 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.