Question

_{（请原谅可怕的人为例子）}

我想要做的是在where子句中指定类型：

somemap :: (a -> b) -> [a] -> [b]
somemap f xs = ys
  where
    some = take 5 xs :: [a]
    ys = map f some :: [b]

但这会导致错误：

*Main> :load file.hs 
[1 of 1] Compiling Main             ( file.hs, interpreted )

fil.hs:15:18:
    Couldn't match expected type `a1' against inferred type `a'
      `a1' is a rigid type variable bound by
           an expression type signature at file.hs:15:25
      `a' is a rigid type variable bound by
          the type signature for `somemap' at file.hs:12:12
      Expected type: [a1]
      Inferred type: [a]
    In the second argument of `take', namely `xs'
    In the expression: take 5 xs :: [a]

file.hs:16:13:
    Couldn't match expected type `b1' against inferred type `b'
      `b1' is a rigid type variable bound by
           an expression type signature at file.hs:16:24
      `b' is a rigid type variable bound by
          the type signature for `somemap' at file.hs:12:17
    In the first argument of `map', namely `f'
    In the expression: map f some :: [b]
    In the definition of `ys': ys = map f some :: [b]
Failed, modules loaded: none.

然而，如果我只是指定具体类型，用Int代替a而Bool代替b，那么没问题：

somemap :: (Int -> Bool) -> [Int] -> [Bool]
somemap f xs = ys
  where
    some = take 5 xs :: [Int]
    ys = map f some :: [Bool]

所以我的问题是：如何在where子句中指定泛型类型和类型约束？

Answer 1

在where子句中，类型变量a和b是 new 类型变量;类型变量没有作用域，因此每个类型的签名都有一个新的类型，就像它们是在顶层定义一样。

如果您打开ScopedTypeVariables扩展名（将{-# LANGUAGE ScopedTypeVariables #-}放在文件顶部），并将somemap的类型声明更改为：

somemap :: forall a b. (a -> b) -> [a] -> [b]

然后您指定的where子句定义将正常工作。我认为forall只是向后兼容性所必需的，因此在where子句中重用多态值的类型变量的代码不会中断。

如果您不想使用扩展名，另一种方法是定义丑陋的帮助函数来统一类型，例如asTypeOf。

如何在“where”子句中赋予表达式泛型类型？

1 个答案: