On 5/31/05, Seth Kurtzberg
Because I'm not smart enough to understand it?
;-)
OK, sorry about the notation (I just didn't feel like doing so much typing, you know :) Here goes... First we observe that, g = new . flip zip [0..], so, without the type specification, it has the general type (New [(a, b1)] b, Num b1, Enum b1) => [a] -> b, as reported by GHC. Then we infer from (1) g :: (New [(u, v)] w, Num v, Enum v) => [u] -> w and (2) instance New [(a, b)] (Map a b) that in (New [(u, v)] w), w can only be (Map u v), so subst (Map u v) for w to obtain (3) g :: (New [(u, v)] (Map u v), Num v, Enum v) => [u] -> Map u v Furthermore, for g we have already specified (4) g :: Ord a => [a] -> Map a Int so finally substituting Int for v should give us (5) g :: (New [(u, Int)] (Map u Int), Num Int, Enum Int, Ord Int) => [u] -> Map u Int Is there any good reason why GHC won't do this? Particularly, is there any good reason not to infer (New [(u, v)] (Map u v)) from (New [(u, v)] w) as in (1), (2) --> (3) above? Cheers, D. Tenev