如何将instanceof与通用函数类型一起使用

Question

在下面的例子中，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会报告以下错误：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类型？

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(在编译过程中被擦除，所以不能说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(这是一个值，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));

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