Re: [Haskell-cafe] Area from [(x,y)] using foldl

On Sun, Nov 8, 2009 at 10:30 PM, michael rice wrote:

This doesn't.

area :: [(Double,Double)] -> Double area p = abs $ (/2) $ area' (last p):p

where area' [] = 0 area' ((x0,y0),(x,y):ps) = ((x0-x)*(y0+y)) + area' (x,y):ps

This function is almost correct except you got your priorities wrong : application priority is always stronger than any operator's so "area' (last p):p" is read as "(area' (last p)) : p"... Besides your second pattern is also wrong, the correct code is : area :: [(Double,Double)] -> Double area p = abs $ (/2) $ area' (last p : p) where area' ((x0,y0):(x,y):ps) = ((x0-x)*(y0+y)) + area' (x,y):ps area' _ = 0 -- Jedaï