在typescript中插入条件类型

Question

很抱歉，对于更熟悉类型系统的人来说，这可能是一个简单的问题。为什么在typescript中内联条件会得到不同的结果？

例如，如果我使用Extract条件，如果我内联它，与将它用作泛型相比，我会得到不同的行为：

// setup
type TUnion = ['foo', 1] | ['baz', 2];
type TDiscriminator = ['foo', any];
// using the conditional generic
type Extract<T, U> = T extends U ? T : never;
type TExtracted = Extract<TUnion, TDiscriminator>; // ["foo", 1]
// inlining the conditional
type TInlined = TUnion extends TDiscriminator ? TUnion : never; // never

当我进一步推动这一点时，似乎需要传递T的值：

type ExtractWithU<T> = T extends TDiscriminator ? T : never;
type TExtractedWithU = ExtractWithU<TUnion>; // ["foo", 1]
type ExtractWithT<U> = TUnion extends U ? TUnion : never;
type TExtractedWithT = ExtractWithT<TDiscriminator>; // never

我是不是错过了一些简单的东西？

Answer 1

从上面带有jcalz和Titian的注释链来看，问题是泛型正在分发，而内联的则没有。

具体为

// setup
type TA = ['foo', 1];
type TB = ['baz', 2];
type TUnion = TA | TB;
type TDiscriminator = ['foo', any];
// using the conditional generic
type Extract<T, U> = T extends U ? T : never; 
type TExtracted = Extract<TUnion, TDiscriminator>; // ["foo", 1]
//              = Extract<TA | TB, TDiscriminator>
//              = Extract<TA, TDiscriminator> | Extract<TB, TDiscriminator>
//              = TA | never
//              = TA
// inlining the conditional
type TInlined = TUnion extends TDiscriminator ? TUnion : never; // never
//            = never, because the union isn't a subtype of the discriminator