[Haskell-cafe] lazy evaluation

[Peter Padawitz]

Can anybody give me a simple explanation why the second definition of a 
palindrome checker does not terminate, although the first one does?

pal :: Eq a => [a] -> Bool
pal s = b where (b,r) = eqrev s r []

eqrev :: Eq a => [a] -> [a] -> [a] -> (Bool,[a])
eqrev (x:s1) ~(y:s2) acc = (x==y&&b,r) where (b,r) = eqrev s1 s2 (x:acc)
eqrev _ _ acc                  = (True,acc)

pal :: Eq a => [a] -> Bool
pal s = b where (b,r) = eqrev' s r

eqrev' :: Eq a => [a] -> [a] -> (Bool,[a])
eqrev' (x:s1) ~(y:s2) = (x==y&&b,r++[y]) where (b,r) = eqrev' s1 s2
eqrev' _ _                   = (True,[])

Peter