无法与实际类型“[a]”匹配的预期类型“a” [英] Couldn't match expected type `a' with actual type `[a]'

问题描述

我能够完美地执行以下代码：

  myLast :: [a]  - > a 
 myLast [] =错误不能在空列表上调用myLast！ 
 myLast（x：_）= x 
  

但是我得到这个错误无法与实际类型[a]匹配的预期类型a。 `a'是一个刚性类型变量，由myLast的类型签名绑定:: [a] - >一个为下面的代码：

  myLast :: [a]  - > a 
 myLast [] =错误不能在空列表上调用myLast！ 
 myLast（_：x）= x 
  

我是Haskell的初学者，错误消息太希腊和拉丁对我来说。根据我的理解，编译器无法在第二种情况下推断出类型。有人可以指点我到底发生了什么吗？

c $ c> [a] ，其余为 a 。

Haskell中的类型 a 的列表由 a 类型的头部和一个尾部，一个类型列表并[a] 。 cons构造函数：以头部和尾部为参数。

当您将列表解构为（x：y）， x 是头部， y 是尾巴。所以在你的第二个代码片段中，当你的类型签名时，你绑定了列表类型的 tail ，列表类型为 [a] 要求你返回一个 a 类型的值（头部是​​一个例子）。

I was able to execute the following code flawlessly

myLast :: [a] -> a
myLast [] = error "Can't call myLast on an empty list!"
myLast (x:_) = x

but I'm getting this error Couldn't match expected type `a' with actual type `[a]'. `a' is a rigid type variable bound by the type signature for myLast :: [a] -> a for the following code:

myLast :: [a] -> a
myLast [] = error "Can't call myLast on an empty list!"
myLast (_:x) = x

I'm a beginner in Haskell and the error message is too greek and latin for me. From what I can understand, the compiler is not able infer the type in the second case. Can someone point me to what is actually happening here?

解决方案

You're declaring the input to be a list of type [a], and the rest to be of type a.

A list of type a in Haskell consists of a head of type a and a tail, a list of type [a]. The cons constructor : takes the head and tail as its arguments.

When you deconstruct a list as (x:y), x is the head and y is the tail. So in your second code fragment, you're binding the tail of the list, which has the list type [a], when your type signature requires that you return a value of type a (the head being one example).

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