函数的类型推断作为参数 [英] Type inference of functions as arguments

问题描述

我想编写一个函数，该函数将几个tupple作为参数，并选择它们的第ith个元素，然后传递给另一个函数，其中i作为另一个参数给出.我已经尝试过这样的事情:

I would like to write a function which takes several tupples as arguments and choose theirs ith elements and passes to another function, where i is given as another argument. I've tried sth like this:

let function (tup1:'A*'A) (tup2:'B*'B) i =
    otherFunction (i tup1) (i tup2)
function Tup1 Tup2 fst

我遇到了一个错误，因为预期i应该是'A*'A ->'A而不是'B*'B->'B. 有什么办法可以使此代码正常工作?

I've got an error, because i was expected to be 'A*'A ->'A not 'B*'B->'B. Is it any way to make this code to work?

谢谢.

推荐答案

您基本上想传递类型为∀'a.'a*'a->'a的参数，但是在F#(和其他ML)中，只有

You basically want to pass an argument of type ∀'a.'a*'a->'a, but in F# (and other MLs), only rank-1 polymorphism is supported so you can't do this directly. The workaround is to define a new type with a generic method to emulate higher-rank polymorphism:

type Untupler =
    abstract Apply : 'a*'a -> 'a

let myFunction tup1 tup2 (i:Untupler) =
    otherFunction (i.Apply tup1) (i.Apply tup2)

myFunction Tup1 Tup2 { new Untupler with member __.Apply (x,y) = x }

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