I'm new to Haskell and functional programming; have been exploring the fixed-point combinator based on the exercise in the HSOE book. Having an odd issue with typing in ghci which I don't understand - maybe it's to do with section 3.4.5. "Type defaulting in GHCi" but I don't really grok that yet... (BTW, an unrelated question: is there a good reference where I can understand the precendence/associativity rules for Haskell? I continually find myself needing lots of parens before expressions behave as I expect, and get very confused trying to use the $ operator - usually resorting to lots of parens again :-) My typing question might boil down to this simple example, though my original example follows: -- Why don't these (particularly g and g') all have the same type? Prelude> :t (\x -> x+1) (\x -> x+1) :: (Num a) => a -> a Prelude> let g = (\x -> x+1) Prelude> :t g g :: Integer -> Integer Prelude> let g' x = x + 1 Prelude> :t g' g' :: (Num a) => a -> a -- Here's my original fixed-point combinator example: Prelude> let fix f = f (fix f) -- here's a silly (but working) implementation of length using fix: Prelude> fix (\g xs -> if xs == [] then 0 else (1 + g (tail xs))) [1,2,3,4] 4 -- so I examine the types of the parts, which seems fine: Prelude> :t fix (\g xs -> if xs == [] then 0 else (1 + g (tail xs))) ... :: (Num t, Eq a) => [a] -> t Prelude> :t (\g xs -> if xs == [] then 0 else (1 + g (tail xs))) ... :: (Num t, Eq a) => ([a] -> t) -> [a] -> t -- but now I try to bind the anonymous function to a name -- this seems to get the types wrong and no longer works as I expect: Prelude> let lenstep = (\g xs -> if xs == [] then 0 else (1 + g (tail xs))) Prelude> :t lenstep lenstep :: ([()] -> Integer) -> [()] -> Integer Prelude> :t fix lenstep fix lenstep :: [()] -> Integer Prelude> let len' = fix (\g xs -> if xs == [] then 0 else (1 + g (tail xs))) Prelude> :t len' len' :: [()] -> Integer Prelude> Prelude> len' [1,2,3,4] <interactive>:1:12: No instance for (Num ()) arising from the literal `4' at <interactive>:1:12 Possible fix: add an instance declaration for (Num ()) In the expression: 4 In the first argument of `len'', namely `[1, 2, 3, 4]' In the expression: len' [1, 2, 3, 4] -- maybe this is just me not understanding name binding properly; it seems to work if I do it this way: Prelude> let lenstep' g xs = (if xs == [] then 0 else (1 + g (tail xs))) Prelude> :t lenstep' lenstep' :: (Eq a, Num t) => ([a] -> t) -> [a] -> t Prelude> :t fix lenstep' fix lenstep' :: (Eq a, Num t) => [a] -> t Prelude> fix lenstep' [1,2,3,4] 4 -- but what's the difference? Cheers, Patrick Patrick.Surry@portraitsoftware.com mailto:Patrick.Surry@portraitsoftware.com , VP Technology Tel: (617) 457 5230 Mob: (857) 919 1700 Fax: (617) 457 5299 Map <http://maps.google.com/maps?f=q&hl=en&q=125+Summer+St,+Boston,+MA+02110 &ie=UTF8&z=15&ll=42.353153,-71.057296&spn=0.022644,0.054245&om=1&iwloc=A

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