Is it possible to eta (or otherwise) reduce arguments if I depend on them?

Question

I was solving a super simple problem and came out with this code. I feel like there should be a way to eliminate arguments x and y in areaOrPerimeter. But I can't find a way to do it. Is it possible to get rid of them somehow? I have a strong feeling that it should be possible.

areaOrPerimeter :: Double -> Double -> Double
areaOrPerimeter x y
  | x == y = area x y 
  | otherwise = perimeter x y
  

area :: Num a => a -> a -> a
area = (*)

perimeter :: Num a => a -> a -> a
perimeter x y = 2 * (x + y)

Answer 1

如果您想有机会以一种不令人讨厌的方式做到这一点，您应该首先将参数有意义地组合在一起，以避免处理多个参数：

data RectangleSize a = RectangleSize {width, height :: a}

然后我们可以使函数和x==y检查也更有意义：

area :: Num a => RectangleSize a -> a
area (RectangleSize w h) = w*h

perimeter :: Num a => RectangleSize a -> a
perimeter (RectangleSize w h) = 2*(w+h)

isSquare :: Eq a => RectangleSize a -> Bool
isSquare (RectangleSize w h) = w==h

这给我们留下了已经更有希望的

areaOrPerimeter :: RectangleSize Double -> Double
areaOrPerimeter r
  | isSquare r  = area r 
  | otherwise   = perimeter r

为了让这一点成为免费，你仍然需要复制三遍r。这可以通过函数实例来完成（是否应该是另一回事） ，即可以写为. 在这种情况下，你想要做出决定，所以你需要一流的；通常是这样写的Applicative\x -> f x (g x)f<*>gif

if' :: Bool -> a -> a -> a
if' True x _ = x
if' False _ y = y

现在您可以isSquare通过组合升级到切换功能if'：

    if' . isSquare :: Eq a => RectangleSize a -> b -> b -> b

现在我们可以使用 applicative 实例将矩形分配给所有组件：

areaOrPerimeter = if'.isSquare <*> area <*> perimeter

在我看来，实际上并不算太破烂。