7

（请原谅这个可怕的人为的例子）

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.

而如果我只是指定具体类型，替换Int为a和Bool，b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？

4

1 回答 1

15

在where子句内部，类型变量a和b是新的类型变量；类型变量没有作用域，因此每个类型签名都有新的供应，就像它们是在顶层定义的一样。

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

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

那么where您指定的子句定义将正常工作。我认为foralls 仅用于向后兼容，因此在where子句中为多态值重用类型变量的代码不会中断。

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

于 2012-05-03T15:18:51.933 回答