无法将预期类型与推断类型，刚性类型变量错误进行匹配

问题描述

此功能有什么问题?

test :: Show s => s
test = "asdasd"

String是Show类的实例，因此似乎正确.

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

错误是

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 => a的意思是对于任何作为Foo实例的类型，test都是该类型的值".因此，在可以使用X类型的值(其中X是实例Foo)的任何地方，都可以使用类型Foo a => a的值.

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.

之所以可以使用test :: Num a => a; test = 42之类的东西，是因为42可以是Int或Integer或Float类型的值，或者是Num实例的其他任何值.

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.

但是"asdasd"不能是Int或作为Show实例的任何其他内容-它只能是String.因此，它与Show s => s类型不匹配.

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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