haskell - Finding maximum element in a list of tuples -

i complete beginner in haskell. have list of tuples i'm using in haskell: structure [(a,b),(c,d),(e,f),(g,h)]

what want return maximum element in tuple according second value: if list of tuples [(4,8),(9,10),(15,16),(10,4)], want maximum element (15,16).

but have no idea how this. attempt far,

maximum' ::  (ord a) => (num a) => [(a,b)] ->   maximum' [] = error "maximum of empty list"   maximum' [(x,y)] = -1 maximum' (x:xs)      | snd x > snd(xs !! maxtail) = 0   | otherwise = maxtail     maxtail = maximum' xs + 1 

and error message makes no sense me:

newjo.hs:23:25: not deduce (a ~ int) context (ord a, num a)   bound type signature              maximum' :: (ord a, num a) => [(a, b)] ->   @ newjo.hs:19:14-47   `a' rigid type variable bound       type signature maximum' :: (ord a, num a) => [(a, b)] ->       @ newjo.hs:19:14 in second argument of `(!!)', namely `maxtail' in first argument of `snd', namely `(xs !! maxtail)' in second argument of `(>)', namely `snd (xs !! maxtail)'` 

i need on how this.

the solutions presented far have been elegant, , should use them in real code write. here's version uses same pattern-matching style you're using.

maximum' :: ord => [(t, a)] -> (t, a) maximum' []     = error "maximum of empty list" maximum' (x:xs) = maxtail x xs   maxtail currentmax [] = currentmax         maxtail (m, n) (p:ps)           | n < (snd p) = maxtail p ps           | otherwise   = maxtail (m, n) ps 

this solution avoids juggling indices around, , instead keeps track of current maximum element, returned when entire list has been traversed. avoiding indices lists considered practice.