"foldl1(\ ab - >(snd a + snd b))[(1,2),(3,4)]的Haskell错误

Question

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)))'

Answer 1

foldl1函数参数必须具有类型a -> a -> a,即2个输入参数,返回值必须具有相同的类型.在你的表达式中,该函数应该返回一个2元组Num b => (a, b),而不是纯数字Num b => b,因此"发生检查".

您可以使用foldl并提供初始值,例如

foldl (\acc elm -> acc + snd elm) 0 [(1,2),(3,4)]

或使用现有的功能

(sum . map snd) [(1,2),(3,4)]