为什么这不起作用？
Prelude> foldl1 (\a b -> ((snd a) + (snd b))) [(1,2),(3,4)]
<interactive>:1:17:
Occurs check: cannot construct the infinite type: b = (a, b)
Expected type: (a, b)
Inferred type: b
In the expression: ((snd a) + (snd b))
In the first argument of `foldl1', namely
`(\ a b -> ((snd a) + (snd b)))'
haskell ×1