Question

考虑以下代码......

type TypeOne () =
    member val Name = "" with get, set

type TypeTwo () =
    member val Name = "" with get, set
    member val Property = 0 with get, set

[<RequireQualifiedAccess>]
type UnionType =
    | TypeOne of TypeOne
    | TypeTwo of TypeTwo

// this only succeeds because we have defined the union type a
// requiring qualified access. otherwise TypeOne would be inferred
// as the union case, not the type (understandably)...

let t1 = TypeOne ()

// if we want t1 "as" the union type, we have to do the following...

let ut1 = UnionType.TypeOne t1

// the following function returns a result of type UnionType
// but we have to explicitly do this for each type

let makeUnion switch =
    match switch with
        | true -> UnionType.TypeOne (TypeOne ())
        | _ -> UnionType.TypeTwo (TypeTwo ())

在评论中，似乎没有办法推断返回结果应该是联合类型，如果我们必须要求对联合类型进行限定访问，这非常冗长（这看起来很令人担忧）。 / p>

也无法创建仅接受union类型的函数并将它们作为union返回。（在这种情况下，一个函数接受TypeOne或TypeTwo并返回UnionType）。还是有吗？是否有更好的方式与受歧视的工会合作？

Answer 1

type TypeOne() =
    member val Name = "" with get, set

type TypeTwo() =
    member val Name = "" with get, set
    member val Property = 0 with get, set

type UnionType =
    | TypeOne of TypeOne
    | TypeTwo of TypeTwo

/// You can give some hints to the type checker 
/// by saying that you want a class constructor
let t1 = new TypeOne()

/// No need to use fully-qualified access on union types
let ut1 = TypeOne t1

let makeUnion switch =
    match switch with
    | true -> TypeOne (new TypeOne())
    | false -> TypeTwo (new TypeTwo())

Answer 2

正如pad指出的那样 - 你不需要用它的父类型限定一个联合案例 - 所以你可以写

let makeUnion switch =
    match switch with
    | true -> TypeOne (new TypeOne ())
    | false -> TypeTwo (new TypeTwo ())

但是，您必须始终指定TypeOne。这是为了避免在存在多个采用相同参数的情况时存在歧义 - 比如

type foo=
|Bar of int
|Baz of int

然后即使知道返回类型，编译器也无法确定返回的内容。一个更常见的例子实际上是不带参数的情况 - 比如说你决定重新划分真假：

type bool = |True |False

理解对受歧视的工会的推断

2 个答案: