从泛型中传递的类型获取枚举对象

这种缩短的编码真正给您带来了什么好处?

IMO,TypeScript中的枚举并不值得。

解决这个问题的方法是使用类型(记住,类型可以是字符串),而不是枚举。这个 Set 类免费为您提供重复删除功能。

class Flags3<T extends string>{
    s: Set<string> = new Set();
    constructor(enu: readonly T[]) {
        // check here the values in enu, fine.
    }
    set(v: T) {
        this.s.add(v)
        return this;
    }
    has(v: T) {
        return this.s.has(v);
    }
}

// parameter type below is ("legged" | "winged")[], NOT string[]
const f3 = new Flags3(["legged", "winged"]);

// or
// const animalTypes=["legged","winged"] as const;
// const f3 = new Flags3(animalTypes)

f3.set("legged"); // OK
f3.set("foo"); // compile time error

通过指定 readonly T[] 在构造函数参数中,我们能够推断传递给它的元组中的各个类型。

Playground Link

Further reading about the downsides of enum

enum TypeAnimal{
    Winged = "w",
    Legged = "l"
}

// modified class from the question
class Flags2<ENU>{
    s: string = "";
    set(v: ENU){
        this.s += v;    // #todo avoid duplicates in this.s
        return this;
    }
    has(v: ENU){
        return this.s.includes(`${v}`);
    }
}

// version with a Set instead of a string. No duplicates.
class Flags3<ENU>{
    s: Set<ENU> = new Set();
    set(v: ENU){
        this.s.add(v);
        return this;
    }
    has(v: ENU){
        return this.s.has(v);
    }
}

let flagsBird2 = new Flags2<TypeAnimal>();
let flagsBird3 = new Flags3<TypeAnimal>();