Question

我只是好奇者，有没有一种方法可以区分原子类型以提高TypeScript的类型安全性？

换句话说，下面是否可以复制行为：

export type Kilos<T> = T & { discriminator: Kilos<T> };   // or something else  
export type Pounds<T> = T & { discriminator: Pounds<T> }; // or something else

export interface MetricWeight {
    value: Kilos<number>
}

export interface ImperialWeight {
    value: Pounds<number>
}

const wm: MetricWeight = { value: 0 as Kilos<number> }
const wi: ImperialWeight = { value: 0 as Pounds<number> }

wm.value = wi.value;                  // Should give compiler error
wi.value = wi.value * 2;              // Shouldn't error, but it's ok if it would, because it would require type casting which asks for additional attention
wm.value = wi.value * 2;              // Already errors
const we: MetricWeight = { value: 0 } // Already errors

或者可以将其放入一个容器的东西：

export type Discriminate<T> = ...

export type Kilos<T> = Discriminate<Kilos<T>>;
export type Pounds<T> = Discriminate<Pounds<T>>;

...

修改

好吧，事实证明，可以使用ZpdDG4gta在此处https://github.com/microsoft/TypeScript/issues/202发现的不可能的类型hack来构建这种类型。

但是当前语言版本有点混乱：

export type Kilos<T> = T & { discriminator: any extends infer O | any ? O : never };
export type Pounds<T> = T & { discriminator: any extends infer O | any ? O : never };

export interface MetricWeight {
    value: Kilos<number>
}

export interface ImperialWeight {
    value: Pounds<number>
}

const wm: MetricWeight = { value: 0 as Kilos<number> }
const wi: ImperialWeight = { value: 0 as Pounds<number> }

wm.value = wi.value;                       // Errors, good
wi.value = wi.value * 2;                   // Errors, but it's +/- ok
wi.value = wi.value * 2 as Pounds<number>; // Shouldn't error, good
wm.value = wi.value * 2;                   // Errors, good
const we: MetricWeight = { value: 0 }      // Errors, good

不幸的是，以下操作无效：

export type Discriminator<T> = T & { discriminator: any extends infer O | any ? O : never } 

export type Kilos<T> = Discriminator<T>;
export type Pounds<T> = Discriminator<T>;

export interface MetricWeight {
    value: Kilos<number>
}

export interface ImperialWeight {
    value: Pounds<number>
}

const wm: MetricWeight = { value: 0 as Kilos<number> }
const wi: ImperialWeight = { value: 0 as Pounds<number> }

wm.value = wi.value;                       // Doesn't error, this is bad
wi.value = wi.value * 2;                   // Errors, but it's +/- ok
wi.value = wi.value * 2 as Pounds<number>; // Shouldn't error, good
wm.value = wi.value * 2;                   // Errors, good
const we: MetricWeight = { value: 0 }      // Errors, good

修改

事实证明，按照@jcalz，还有另一种方法来引入不可能的类型：

export type Kilos<T> = T & { readonly discriminator: unique symbol };
export type Pounds<T> = T & { readonly discriminator: unique symbol };

...

但是仍然缺少

export type Discriminator<T> = ...

有什么想法使其变得更清洁？由于类型别名使两个类型引用都坚持使用Discriminator ...

Answer 1

只需将其定义为：

Save

然后，磅和公斤会自动投射到数字和数字上，而不是彼此投射。

const marker = Symbol();

export type Kilos = number & { [marker]?: 'kilos' };
export const Kilos = (value = 0) => value as Kilos;

export type Pounds = number & { [marker]?: 'pounds' };
export const Pounds = (value = 0) => value as Pounds;

如果要防止自动将“增强”单位转换为“普通”数字，则只需从标记字段中删除可选内容即可：

let kilos = Kilos(0);
let pounds = Pounds(0);
let wrong: Pounds = Kilos(20); // Error: Type 'Kilos' is not assignable to type 'Pounds'.

kilos = 10; // OK
pounds = 20;  // OK

let kilos2 = 20 as Kilos; // OK
let kilos3: Kilos = 30; // OK

pounds = kilos;  // Error: Type 'Kilos' is not assignable to type 'Pounds'.
kilos = pounds; // Error: Type 'Pounds' is not assignable to type 'Kilos'.

kilos = Kilos(pounds / 2); // OK
pounds = Pounds(kilos * 2); // OK

kilos = Pounds(pounds / 2); // Error: Type 'Pounds' is not assignable to type 'Kilos'.

kilos = pounds / 2; // OK
pounds = kilos * 2; // OK

TypeScript中的原子类型判别（标称原子类型）

1 个答案: