Question

sumOfSquare :: Int -> Int -> Int
sumOfSquare a b = a * a + b * b

hipotenuse :: Int -> Int -> Int
hipotenuse a b = truncate(sqrt(x))
           where x = fromIntegral(sumOfSquare a b)

squareCheck :: Int -> Bool
squareCheck n = truncate(sqrt(x)) * truncate(sqrt(x)) == n
         where x = fromIntegral n

isItSquare :: Int -> Int -> Bool
isItSquare a b = squareCheck (sumOfSquare a b)

calc :: (Integral a) => a -> [(a, a, a)]
calc a = [(x, y, (hipotenuse x y)) | x <- [1..a], y <-[1..a], (isItSquare x y)]

错误讯息：

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.

据我了解'x'和'y'的类型。这样对吗？是方形需要Int。但是'x'和'y'是什么类型的？我认为他们是Int。

Answer 1

你的类型太笼统了。您正在将x和y传递给isItSquare，这是Int，但您不知道x和y是Int秒。它们可以是，但它们也可以是Integral的任何其他实例。将签名更改为更具体的：

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

或者让您的帮助函数适用于更常规的类型：

squareCheck :: (Integral a) => a -> Bool
...

Answer 2

您已将sumOfSquare，hipotenuse，squareCheck和isItSquare声明为Int。

但是，您已经说calc可以使用任何类型a，只要a为Integral。

要么像这样声明calc：

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

...或更改所有其他功能：

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

Answer 3

calc :: (Integral a) => a -> [(a, a, a)]
calc a = [(x, y, (hipotenuse x y)) | x <- [1..a], y <-[1..a], (isItSquare x y)]

a的类型为a（签名明确表示如此，这是“calc的类型签名绑定的刚性类型变量”和{{1} }取自列表x，因此它也有[1..a]类型（a也相同）。

而另一种类型的错误

3 个答案: