Question

在下面的示例中，我想确保返回的类型与泛型函数中给定的泛型类型相对应。不幸的是，在使用result instanceof T时，TypeScript报告以下错误：error TS2693: 'T' only refers to a type, but is being used as a value here。有没有一种方法可以在运行时检查对象的类型是否对应于泛型？

class Foo {
    id: number;
    constructor(id: number) {
        this.id = id;
    }
}

class Bar extends Foo {
    name: string;
    constructor(id: number, name: string) {
        super(id);
        this.name = name;
    }
}

function get<T>(list: Array<Foo>, id: number): T|undefined {
    const result = list.find(e => e.id === id);
    if (typeof result === 'undefined') {
        return undefined;
    }

    if (!(result instanceof T)) { // error TS2693: 'T' only refers to a type, but is being used as a value here 
        return undefined;
    }

    return result as unknown as T;
}

const foos = [new Foo(1), new Bar(2, 'bar'), new Foo(3)];

console.log(get<Foo>(foos, 1));
console.log(get<Foo>(foos, 2));
console.log(get<Bar>(foos, 3));

Answer 1

类型（和类型参数）在编译期间被删除，因此您不能说result instanceof T，因为在运行时没有T。

您可以做的是将类本身传递给函数（它是一个值，因此可以在运行时使用）


function get<T>(cls: new (...a: any) => T, list: Array<Foo>, id: number): T|undefined {
    const result = list.find(e => e.id === id);
    if (typeof result === 'undefined') {
        return undefined;
    }

    if (!(result instanceof cls)) {
        return undefined;
    }

    return result as unknown as T;
}

const foos = [new Foo(1), new Bar(2, 'bar'), new Foo(3)];

console.log(get(Foo, foos, 1));
console.log(get(Foo, foos, 2));
console.log(get(Bar, foos, 3));

Playground Link

如何将instanceof与泛型函数类型一起使用

1 个答案: