刚性类型变量错误 [英] rigid type variable error

问题描述

这个函数有什么问题？

test :: Show s ⇒ s
test = "asdasd"

字符串是Show类的一个实例，所以看起来正确。

String is an instance of Show class, so seems correct.

错误是：

The error is

src\Main.hs:224:7:
    Couldn't match expected type `s' against inferred type `[Char]'
      `s' is a rigid type variable bound by
          the type signature for `test' at src\Main.hs:223:13
    In the expression: "asdasd"
    In the definition of `test': test = "asdasd"

推荐答案

test :: Foo a =>一个的意思是对于任何类型的实体 Foo ， test 是一个值这种类型。因此，在任何可以使用类型 X 的值的地方，其中 X 是一个实例 Foo ，您可以使用类型的值Foo a =>一个。

test :: Foo a => a means "for any type which is an instance of Foo, test is a value of that type". So in any place where you can use a value of type X where X is an instance Foo, you can use a value of type Foo a => a.

类似于 test :: Num a =>一个; test = 42 可以工作，因为42可以是类型 Int 或整数或 Float 或其他任何实例 Num 。

Something like test :: Num a => a; test = 42 works because 42 can be a value of type Int or Integer or Float or anything else that is an instance of Num.

然而asdasd不能是 Int 或其他任何其他实体显示 - 它只能是字符串。因此，它与 Show s =>类型不匹配。 s 。

However "asdasd" can't be an Int or anything else that is an instance of Show - it can only ever be a String. As a consequence it does not match the type Show s => s.

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