为什么Haskell不允许我对其函数定义更加宽松？

Question

我发现这一点有点惊讶

head' :: [a] -> b
head' (x:xs) = x

提出一个

Couldn't match expected type `b' with actual type `a'
  `b' is a rigid type variable bound by
      the type signature for head' :: [a] -> b at type_test.hs:1:10
  `a' is a rigid type variable bound by
      the type signature for head' :: [a] -> b at type_test.hs:1:10
In the expression: x
In an equation for head': head' (x : xs) = x

这是为什么？我假设Haskell会允许我像我想要的那样松懈,并且会发现没问题[a] -> b.

谢谢

Answer 1

功能类型签名不正确.由于输入是类型[a],输出将始终是类型a.类型签名[a] -> b表示该函数将接受任何事物列表,并返回任何(可能是不同的)类型的东西,这是不正确的 - 它只能返回相同类型的东西a.