如何在定义的F＃函数中交换应用程序的顺序？

Given a general function f:a->b->c

给定一般函数f:a-> b-> c

I want to create the function that does the same thing/computation that f does but with the order of the two input parameters (or the currying order) swapped around.

我想创建一个函数,它执行与f相同的事情/计算,但是交换了两个输入参数(或currying命令)的顺序。

So looking for g:b->a->c where g does the exact same computation as f

所以寻找g:b-> a-> c其中g与f完全相同的计算

let transform1 (f:a->b->c) : (b->a->c) = 
?

3 个解决方案

#1

You can actually tell from the type of the function how to do this.

您实际上可以从函数的类型告诉如何执行此操作。

transform1 has the type (a -> b -> c) -> (b -> a -> c) or, equivalently, (a -> b -> c) -> b -> a -> c.

transform1具有类型(a - > b - > c) - >(b - > a - > c)或等效地,(a - > b - > c) - > b - > a - > c。

So what you're looking for is a function that takes

所以你要找的是一个需要的功能

a function a -> b -> c which we'll call f,

函数a - > b - > c我们称之为f,

and a b,

和一个b,

and an a,

和一个,

and then "uses" f with the a and the b in order to produce a c.

然后用a和b“使用”f以产生c。

Since there's only one way to "use" f with the a and the b that produces a c you write that way down:

由于只有一种方法可以“使用”f与a和b产生一个c,你可以这样写下来:

flip f b a = f a b

and that's it.

就是这样。

#2

You can define a swap function thus:

您可以定义交换功能:

let swap f a b = f b a

#3

In Haskell you'd do that like this:

在Haskell你会这样做:

g b a = f a b -- or using lambda expressions: (\b a -> f a b)

So in a let statement: let g = (\b a -> f a b) in ...

所以在一个let语句中:让g =(\ b a - > f a b)in ...

#1