空列表中的 F# 值限制 [英] F# value restriction in empty list

问题描述

我有一个 F# 函数:

I have a F# function:

let removeEven (listToGoUnder : _ list) =
    let rec listRec list x =
        match list with
        | [] -> []
        | head::tail when (x%2 = 0) -> head :: listRec (tail) (x+1)
        | head::tail -> listRec (tail) (x+1)

     listRec listToGoUnder 0

它删除列表中偶数索引处的所有元素.如果我给列表一些输入，它会起作用，比如 removeEven ['1';'2';'3'] 我得到 ['1';'3']我应该这样做.但是当我插入一个空列表作为参数时，我得到这个错误:

It removes all elements at an even index in a list. It works if I give the list some imput, like removeEven ['1';'2';'3'] I get ['1';'3'] which I am supposed to. But when I insert a empty list as parameter, I get this error:

stdin(78,1):错误 FS0030:值限制.它"的价值是推断具有泛型类型

stdin(78,1): error FS0030: Value restriction. The value 'it' has been inferred to have generic type

val it : '_a list 或者将 'it' 定义为一个简单的数据项，使它是一个带有显式参数的函数，或者，如果您不打算使用它要通用，请添加类型注释.

val it : '_a list Either define 'it' as a simple data term, make it a function with explicit arguments or, if you do not intend for it to be generic, add a type annotation.

有人帮忙吗?

推荐答案

空列表([])比较特殊；它可以是任何类型的列表.因此，编译器会抱怨您没有 [] 的特定类型.在参数上添加类型注释有助于解决问题:

The empty list ([]) is quite special; it can be a list of any type. Therefore, the compiler complains that you don't have a specific type for []. Adding type annotation on the argument helps to solve the problem:

let results = removeEven ([]: int list)

或@kvb 建议的更多惯用类型注释:

let results: int list = removeEven []

这可能超出问题范围，但您的函数应该命名为 removeOdd，因为索引通常从 0 开始，并且您的函数会删除所有具有奇数索引的元素.此外，如果您在列表的前两个元素上使用模式匹配，而不是保留一个计数器 x 来检查索引，事情会更加清楚:

This is probably beyond the question, but your function should be named as removeOdd since indices often start from 0 and your function removes all elements with odd indices. Moreover, things are much more clear if you use pattern matching on first two elements of the list rather than keep a counter x for checking indices:

let rec removeOdd = function
    | [] -> []
    | [x] -> [x]
    | x::_::xs -> x::removeOdd xs

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