值限制:已将类型"bar"推断为具有通用类型 [英] Value restriction: The type 'bar' has been inferred to have generic type

问题描述

在下面的代码段中，我看不到为什么必须按功能foo的方式编写f和g以及为何按功能bar的方式无法实现.

In the snippet below, I cannot see why I have to compose f and g the way function foo does and why it does not work the way function bar tries to do it.

let f a b = a,b
let g (a : 'a) (b : 'a) = a

let (>!) f1 f2 =
    fun a b ->
        let (x,y) = f1 a b
        f2 x y

let foo = fun a b -> (f >! g) a b
let bar = f >! g

任何人都可以向我解释为什么bar无法正常工作吗?鉴于foo也具有通用类型，对我来说这没有意义.

Can anyone explain to me, why bar is not working? Given that foo also has generic type, it makes no sense to me.

推荐答案

foo是一个函数，而bar是一个值.是的，它是一个函数类型的值，但仍然是一个值.那里有一个细微的差别.

foo is a function, while bar is a value. Yes, it's a value of a function type, but still a value. There is a subtle difference there.

F#编译器可以看到" foo是一个函数，因为它可以在let之后立即看到fun ->.

The F# compiler can "see" that foo is a function, because it sees the fun -> right after let.

您的bar是真实值-通过调用其他函数(运算符>!)获得的结果. F#有一条规则(称为值限制")(每天)说，值(不同于函数)不能具有泛型类型，除非显式指定了泛型参数，从而有效地使其成为类型函数". (比这稍微复杂一点，请阅读下面的链接以获取完整图片)

Your bar, on the other hand, is a true value - a result obtained by invoking a different function (the operator >!). F# has a rule (known as "value restriction") saying (in everyday terms) that values (unlike functions) cannot have generic types, unless generic arguments are specified explicitly, thus effectively making it a "type function". (it's a bit more complicated than that, read the links below for the full picture)

这不是特定于F#的-ML的其他非纯变体也具有此功能. 此处是对此规则的讨论F#和此处是关于SML的讨论.

This is not specific to F#, though - other non-pure variants of ML have this as well. Here is a discussion of this rule for F#, and here's a discussion for SML.

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