Question

给出界面：

interface Extendable<T> {
    extend<U>(q: U): T & U
}

当T和U的交点为空时，您将如何修改它以便出现编译错误？

即T & U === never。

在我的用例中，类型T和U是特定的字符串值或它们的并集。

下面是如何使用该接口的示例：

type S = 'Select'
type U = 'Update'
type D = 'Delete'
type I = 'Insert'
type V = 'Values'
type SUDI = S | U | D | I
type SUD = S | U | D
type SV = S | V

declare let sud: SUD, s: S, i: I;
declare let root: Extendable<SUDI>
declare let where: Extendable<SUD>

root.extend(sud)
// return type (S | U | D | I) & (S | U | D) = (S | U | D)

where.extend(i)
// return type (S | U | D) & (I) = never
// can this be a compile error instead?

Answer 1

您要表达的关系实际上只是一个扩展关系。如果您有工会，则工会成员的一部分将扩展原始工会

interface Extendable<T> {
    extend<U extends T>(q: U): T & U
}

type S = 'Select'
type U = 'Update'
type D = 'Delete'
type I = 'Insert'
type SUDI = S | U | D | I
type SUD = S | U | D

declare let sud: SUD, s: S, i: I;
declare let root: Extendable<SUDI>
declare let where: Extendable<SUD>

root.extend(sud) //ok
where.extend(i) // I !extends SUD

修改

如果两个联合之间没有公共成员，我们可以使用条件类型来触发错误。由于该关系不是传统的关系，因此该错误也不会是传统的编译器错误。如果不满足条件，我们将额外的字符串文字类型添加到参数中，这将触发错误。

type ValidateCommonunionMemebers<T, U, TErr> = [T] extends [U] ? {} :
    [U] extends [T] ? {} : TErr;

interface Extendable<T> {
    extend<U>(q: U & ValidateCommonunionMemebers<T, U, "T and U should have some common memebers">): T & U
}

type S = 'Select'
type U = 'Update'
type D = 'Delete'
type I = 'Insert'
type V = 'Values'
type SUDI = S | U | D | I
type SUD = S | U | D
type SV = S | V

declare let sud: SUD, s: S, i: I, sudi:SUDI;
declare let root: Extendable<SUDI>
declare let where: Extendable<SUD>

root.extend(sud)
// return type (S | U | D | I) & (S | U | D) = (S | U | D)

where.extend(i) // Error Type '"Insert"' is not assignable to type '"T and U should have some common memebers"'.
where.extend(sudi)

Answer 2

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除非T & U === never或T & U都不是，否则

never永远不会是T。

您不能将U用作泛型的占位符，因为它可以分配给任何对象，例如

never

如何在类型T和U上声明泛型函数，不允许T和U的空交集

2 个答案: