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Typescript-有些省略实现从属性中移除可选性

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Ran Lottem · 技术社区 · 6 年前

Omit 省略 本身。我的理解有点困难 为什么有些实现是同态的,而有些则不是

我发现:

悬停时显示的实现 省略 不是正确的
悬停时显示的实现 省略

这是我的代码:

// a type with optional and readonly properties
type HasOptional = { a: number; b?: number, c: string; d?: readonly string[]; };

// first attempt
type Omit_1<T, K extends keyof T> = { [P in Exclude<keyof T, K>]: T[P]; };
type Omit_1_Optional = Omit_1<HasOptional, 'a'>; // b, d lost optionality

// Omit's 'fake' implementation, as shown by Intellisense
type Omit_2<T, K extends string | number | symbol> = { [P in Exclude<keyof T, K>]: T[P]; };
type Omit_2_Optional = Omit_2<HasOptional, 'a'>; // b, d lost optionality

// Using Omit itself
type Omit_3<T, K extends string | number | symbol> = Omit<T, K>;
type Omit_3_Optional = Omit_3<HasOptional, 'a'>; // optionality maintained!

// Writing Omit's implementation explicitly
type Omit_4<T, K extends keyof any> = Pick<T, Exclude<keyof T, K>>;
type Omit_4_Optional = Omit_4<HasOptional, 'a'>; // optionality maintained!

here, in an answer about deep Omit [P in K]: 被用作额外的间接级别,以导致同态行为,但这也存在于这里,但前两个实现并没有保留“可选性”。

1 回复 | 直到 6 年前

1

Titian Cernicova-Dragomir 6 年前

映射类型在两种情况下被认为是同态的。要么我们画一张地图 keyof T ( docs )或者我们映射一个类型参数 K 哪里有一个约束 T键 ( K extends keyof T , docs

当 Exclude<keyof T, K> 不适用于这两种特定情况。这意味着直接映射到不会生成同态映射类型。如果我们采取并将其放入具有所需约束的类型参数中,然后我们获得所需的行为。

// a type with optional and readonly properties
type HasOptional = { a: number; b?: number, c: string; d?: readonly string[]; };

// mapping over Exclude<keyof T, K> optionality lost
type Omit_1<T, K extends keyof T> = { [P in Exclude<keyof T, K>]: T[P]; };
type Omit_1_Optional = Omit_1<HasOptional, 'a'>; // b, d lost optionality

// mapping over Exclude<keyof T, K> optionality lost
type Omit_2<T, K extends string | number | symbol> = { [P in Exclude<keyof T, K>]: T[P]; };
type Omit_2_Optional = Omit_2<HasOptional, 'a'>; // b, d lost optionality

// Omit in 3.5 has homomorphic behavior since it uses Pick which is  homomorphic 
type Omit_3<T, K extends string | number | symbol> = Omit<T, K>;
type Omit_3_Optional = Omit_3<HasOptional, 'a'>; // optionality maintained!

// has homomorphic behavior since it uses Pick which is  homomorphic
type Omit_4<T, K extends keyof any> = Pick<T, Exclude<keyof T, K>>;
type Omit_4_Optional = Omit_4<HasOptional, 'a'>; // optionality maintained!