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 For email: http://www.ninebynine.org/#Contact