On Sun, Nov 8, 2009 at 10:30 PM, michael rice <nowg...@yahoo.com> 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
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