带有元组参数的Haskell中的函数组合 [英] Function composition in Haskell with tuple arguments

问题描述

有时我有两种形式的函数：

Sometimes I have two functions of the form:

f :: a -> (b1,b2)
h :: b1 -> b2 -> c

并且我需要构图g。我通过将h改为h来解决这个问题：

and I need the composition g. I solve this by changing h to h':

h' :: (b1,b2) -> c

您可以告诉我（如果可能的话）函数m，以便：

Can you please show me (if possible) a function m, so that:

(h . m . f) == (h' . f)

或者另一种处理这种情况的方式。感谢。

Or another way to deal with such situations. Thanks.

推荐答案

你正在做的是采用一个在 curried 参数， h ，并将其应用于元组的结果 f 。这个过程将两个参数的函数转换为一个带有一个元组参数的函数，称为 uncurrying 。我们从 Data.Tuple < a>：

What you're looking to do is to take a function that operates on curried arguments, h, and apply it to the result of f, which is a tuple. This process, turning a function of two arguments into a function that takes one argument that is a tuple, is called uncurrying. We have, from Data.Tuple:

curry :: ((a, b) -> c) -> a -> b -> c 
   -- curry converts an uncurried function to a curried function.

uncurry :: (a -> b -> c) -> (a, b) -> c
   -- uncurry converts a curried function to a function on pairs.

现在我们可以写出：

So now we can write:

f :: a -> (b,c)
f = undefined

h :: b -> c -> d
h = undefined

k :: a -> d
k = uncurry h . f

---

另一个难以想象的方法是通过应用仿函数b
$ b

---

Another tricky way to think of this is via an applicative functor,

k = (h <$> fst <*> snd) . f

来自Conor McBride的想法，他会把它写成： f fst snd |）。 f 我想。

Idea from Conor McBride, who'd write it as: (|f fst snd|) . f I think.

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