The type of foldl is:
(b -> a -> b) -> b -> [a] -> b
What do you expect 'a' and 'b' to be in your algorithm?
2009/11/8 michael rice
Here's an (Fortran) algorithm for calculating an area, given one dimensional arrays of Xs and Ys. I wrote a recursive Haskell function that works, and one using FOLDL that doesn't. Why would Haskell be "expecting" (t, t) out of ((*) (xold-x) (yold+y))?
Michael
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AREA = 0.0 XOLD = XVERT(NVERT) YOLD = YVERT(NVERT) DO 10 N = 1, NVERT X = XVERT(N) Y = YVERT(N) AREA = AREA + (XOLD - X)*(YOLD + Y) XOLD = X YOLD = Y 10 CONTINUE
AREA = 0.5*AREA
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area :: [(Double,Double)] -> Double area ps = abs $ (/2) $ area' (last ps) ps where area' _ [] = 0 area' (x0,y0) ((x,y):ps) = (x0-x)*(y0+y) + area' (x,y) ps
*Main> let p = [(0.0,0.0),(1.0,0.0),(1.0,1.0),(0.0,1.0),(0.0,0.0)] *Main> area (last p) p 1.0 *Main>
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area :: [(Double,Double)] -> Double area p = foldl (\ (xold,yold) (x,y) -> ((*) (xold-x) (yold+y))) 0 ((last p):p)
Prelude> :l area [1 of 1] Compiling Main ( area.hs, interpreted )
area.hs:29:40: Occurs check: cannot construct the infinite type: t = (t, t) Expected type: (t, t) Inferred type: t In the expression: ((*) (xold - x) (yold + y)) In the first argument of `foldl', namely `(\ (xold, yold) (x, y) -> ((*) (xold - x) (yold + y)))' Failed, modules loaded: none. Prelude>
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