Haskell 还有一个类型错误

正如我所理解的“x”和“y”的类型。是这样吗？它是正方形吗？需要整数。但是“x”和“y”的类型是什么？我想它们是Int。

您已经声明

sumOfSquare

，

hipotenuse

，

squareCheck

和

isItSquare

在

Int

上运行

但是，您已经说过，

calc

可以使用任何类型的

a

，只要

a

是

整数

像这样声明

calc

：

Prelude> :load "some.hs"
[1 of 1] Compiling Main             ( some.hs, interpreted )

some.hs:16:74:
    Couldn't match expected type `Int' against inferred type `a'
      `a' is a rigid type variable bound by
          the type signature for `calc' at some.hs:15:18
    In the first argument of `isItSquare', namely `x'
    In the expression: (isItSquare x y)
    In a stmt of a list comprehension: (isItSquare x y)
Failed, modules loaded: none.

calc :: Int -> [(Int, Int, Int)]

…或更改所有其他功能，如下所示：

Prelude> :load "some.hs"
[1 of 1] Compiling Main             ( some.hs, interpreted )

some.hs:16:74:
    Couldn't match expected type `Int' against inferred type `a'
      `a' is a rigid type variable bound by
          the type signature for `calc' at some.hs:15:18
    In the first argument of `isItSquare', namely `x'
    In the expression: (isItSquare x y)
    In a stmt of a list comprehension: (isItSquare x y)
Failed, modules loaded: none.

calc :: Int -> [(Int, Int, Int)]

---

您已声明

sumOfSquare

、

hipotenuse

、

squareCheck

和

isItSquare

在

Int

上运行

但是，您已经说过，

calc

可以使用任何类型的

a

，只要

a

是

整数

像这样声明

calc

：

Prelude> :load "some.hs"
[1 of 1] Compiling Main             ( some.hs, interpreted )

some.hs:16:74:
    Couldn't match expected type `Int' against inferred type `a'
      `a' is a rigid type variable bound by
          the type signature for `calc' at some.hs:15:18
    In the first argument of `isItSquare', namely `x'
    In the expression: (isItSquare x y)
    In a stmt of a list comprehension: (isItSquare x y)
Failed, modules loaded: none.

calc :: Int -> [(Int, Int, Int)]

…或更改所有其他功能，如下所示：

Prelude> :load "some.hs"
[1 of 1] Compiling Main             ( some.hs, interpreted )

some.hs:16:74:
    Couldn't match expected type `Int' against inferred type `a'
      `a' is a rigid type variable bound by
          the type signature for `calc' at some.hs:15:18
    In the first argument of `isItSquare', namely `x'
    In the expression: (isItSquare x y)
    In a stmt of a list comprehension: (isItSquare x y)
Failed, modules loaded: none.

calc :: Int -> [(Int, Int, Int)]

---

你的类型太笼统了。您正在将

x

和

y

传递到

isItSquare

，这将增加

Int

s，但您不知道

x

和

y

是

Int

s。它们可以是，但也可以是

Integral

的任何其他实例。将签名更改为更具体的：

sumOfSquare :: (Integral a) => a -> a -> a

或者让助手函数处理更一般的类型：

calc :: Int -> [(Int, Int, Int)]

---

你的类型太笼统了。您正在将

x

和

y

传递到

isItSquare

，这将增加

Int

s，但您不知道

x

和

y

是

Int

s。它们可以是，但也可以是

Integral

的任何其他实例。将签名更改为更具体的：

sumOfSquare :: (Integral a) => a -> a -> a

或者让助手函数处理更一般的类型：

calc :: Int -> [(Int, Int, Int)]

calc:：（积分a）=>a->[（a，a，a）]
计算a=[（x，y，（hipotenuse x y））|x
calc:：（积分a=>a->[（a，a，a）]
计算a=[（x，y，（hipotenuse x y））|x