无法正确理解打字稿中的`Exclude<T, U>` [英] Can't understand `Exclude<T, U>` in typescript correctly
问题描述
打字稿中有类型定义:
type Exclude<T, U> = T extends U ? never : T;
我们可以用它从另一个类型中排除一个类型:
We can use it to exclude a type from another type:
type AB = 'a' | 'b'
type AC = 'a' | 'c'
type X = Exclude<AB, AC>
类型 X 现在是 b.
但是当我直接使用Exclude的内容时:
But when I use the content of Exclude directly:
type X = AB extends AC ? never : AC;
类型X不同，不再是b，而是AC.
The type X is different, it's not b anymore, it's AC.
我不明白为什么它的行为不同.
I can't understand why it behaves differently.
推荐答案
首先，如果替换内容，X 将是 type X = AB extends AC ?从不:AB;.但这也不会给你相同的结果.
Firstly if you replace the content, X would be type X = AB extends AC ? never : AB;. But that will not give you the same result either.
原因是条件类型在涉及裸类型参数时有一种特殊的行为，如果它们是联合，它们会分布在它们之上.因此，当您使用 Exclude 时，联合的每个成员都单独通过条件并合并结果.所以 Exclude 等价于:
The reason is that conditional types have a special behavior when it comes to naked type parameters, they distribute over them if they are a union. So when you use Exclude, each member of the union is separately put through the condition and the result is unioned. So Exclude<AC, AB> is equivalent to:
type X = ('a' extends AC ? never : 'a') | ('b' extends AC ? never : 'b')
除了裸类型参数(裸意味着 T 不用于另一种类型，例如元组、数组或作为另一个泛型类型的参数)外，不会发生这种分配行为，这就是直接在条件中使用类型不会产生相同结果的原因.您可以在此处
This distributive behavior does not occur with anything except naked type parameters (naked meaning T is not used in another type, such as a tuple, an array or as a parameter to another generic type), this is why using the type directly in the condition does not yield the same result. You can read more here
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