如何理解和修复“无法匹配类型”错误？

Question

我正在尝试制作一个 function （ conta conta:: Int -> Polinomio -> Int ），它给了我一个（ conta np ）告诉我我在p中有多少n 。

type Polinomio = [Monomio]
type Monomio = (Float,Int)

conta :: Int -> Polinomio -> Int
conta n [] = 0
conta n ((x,y):xs) = if n == y then x else conta xs

 Ficha2.hs:83:37: error: • Couldn't match expected type 'Int' with actual type 'Float' • In the expression: x In the expression: if n == y then x else conta xs In an equation for 'conta': conta n ((x, y): xs) = if n == y then x else conta xs Ficha2.hs:83:44: error: • Couldn't match expected type 'Int' with actual type 'Polinomio -> Int' • Probable cause: 'conta' is applied to too few arguments In the expression: conta xs In the expression: if n == y then x else conta xs In an equation for 'conta': conta n ((x, y): xs) = if n == y then x else conta xs Ficha2.hs:83:50: error: • Couldn't match expected type 'Int' with actual type '[Monomio]' • In the first argument of 'conta', namely 'xs' In the expression: conta xs In the expression: if n == y then x else conta xs

Answer 1

两个错误：

1. conta xs应该是conta n xs 。 递归调用需要与初始调用相同类型的 arguments。
2. conta 的类型应该是Int -> Polinomio -> Float ，而不是Int -> Polinomio -> Int 。 这是因为它返回元组的第一部分，即Float 。

Answer 2

阅读@Josepf Sible 的出色答案，您应该做的更正是：

type Polinomio = [Monomio]
type Monomio = (Float,Int)

conta :: Int -> Polinomio -> Float
conta n [] = 0
conta n ((x,y):xs) = if n == y then x else conta n xs

例子：

 conta 3 [(4.4,5),(5.2,3)]
=> 5.2

Answer 3

我们来看看错误：

 Ficha2.hs:83:37: error: • Couldn't match expected type 'Int' with actual type 'Float' • In the expression: x In the expression: if n == y then x else conta xs In an equation for 'conta': conta n ((x, y): xs) = if n == y then x else conta xs

x有问题

if n == y then x else conta xs

问题是x是（实际类型） Float ，而实际上它应该是Int 。 它应该是一个Int ，因为您承诺从conta返回一个Int而x是conta返回的值：

conta :: Int -> Polinomio -> Int

它是一个Float ，因为您使用((x, y): xs)来匹配Polinomio并最终匹配(Float, Int) 。 因此我们从conta的第二个参数的类型知道x是一个Float 。

type Polinomio = [Monomio]
type Monomio = (Float, Int)
conta :: Int -> Polinomio -> Int
conta n ((x, y) : xs) = ...