Why does value restriction happen with MergeSort function?

问题

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)

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

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

I encountered a value restriction error:

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.

Why does it happen? What is a simple way to fix it?

回答1:

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).

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)

来源：https://stackoverflow.com/questions/12671112/why-does-value-restriction-happen-with-mergesort-function

标签

mergesort

value-restriction