为什么MergeSort函数会发生值限制? [英] Why does value restriction happen with MergeSort function?

问题描述

我在List上有一个非常简单的 MergeSort 实现.

I have a very simple MergeSort implementation on List.

/// Divide the list into (almost) equal halves
let rec split = function
    | [] -> [], []
    | [x] -> [x], []
    | x1::x2::xs -> let xs1, xs2 = split xs
                    x1::xs1, x2::xs2

/// Merge two sorted lists
let rec merge xs ys =
    match xs, ys with
    | [], _ -> ys
    | _, [] -> xs
    | x::xs', y::ys' when x <= y -> x::merge xs' ys
    | _, y::ys' -> y::merge xs ys' 

let rec mergeSort = function
    | [] -> []
    | xs -> let xs1, xs2 = split xs
            merge (mergeSort xs1) (mergeSort xs2)

但是只要我尝试使用F#Interactive中的任何输入进行测试:

But whenever I tried to test with any input in F# Interactive:

let xs = mergeSort [1;4;3;2];;

我遇到了值限制错误:

错误FS0030:值限制.值"xs"已推断为 具有通用类型 val xs:'_a列出'_a:比较时要么将'xs'定义为一个简单的数据项，使其成为具有显式参数的函数，或者如果 您不希望它具有通用性，请添加类型注释.

error FS0030: Value restriction. The value 'xs' has been inferred to have generic type val xs : '_a list when '_a : comparison Either define 'xs' 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.

为什么会发生?修复它的简单方法是什么?

推荐答案

您没有处理mergeSort中1元素列表的特殊情况. 一般情况是太笼统"，无法推断出正确的类型.因此，编译器会为函数推断出过于通用的类型('a列表->'b列表)，结果始终是通用列表(由于值限制而被禁止).

You are not handling the special case of 1-element lists in mergeSort. The general case is "too general" to infer the right type. As a consequence, the compiler infers a too generic type for the function ('a list -> 'b list) and the result is always a generic list (which is not allowed due to value restriction).

如果您这样修复它，则该类型将正确地推断为列表->列表".

If you fix it like this, the type will be correctly inferred as 'a list -> 'a list.

let rec mergeSort = function
    | [] -> []
    | [x] -> [x]
    | xs -> let xs1, xs2 = split xs
            merge (mergeSort xs1) (mergeSort xs2)

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