Question

我有一个基类（Base）和两个子类A和B.

Base有一个方法f，它创建当前类的实例并返回它。如何确保在调用A.f（）时，它返回A而不是Base的实例，对B来说同样如此？我不想在A或B上定义这个方法。

以下是一些示例代码：

class Base {
    name: string;

    constructor(name: string) { this.name = name; }

    f(): Base /* ? */ {
        return Base(name); /* or A(name) or B(name) depending on what "this" is

    }
}

class A extends Base { /* stuff */ }

class B extends Base { /* stuff */ }

var a = new A();
a.f() -> returns an instance of A, not Base

Answer 1

您可以利用constructor属性：

class Base {
    name: string;

    constructor(name: string) { this.name = name; }

    f(name): Base {
        var c:any = this.constructor;
        return new c(name); /* or A(name) or B(name) depending on what "this" is */
    }
}

class A extends Base { /* stuff */ }

class B extends Base { /* stuff */ }

var a = new A("something");
var foo = a.f("foo") // -> returns an instance of A, not Base
console.log(foo.constructor);  // "A"

var b = new B("something else");
var bar = b.f("bar");
console.log(bar.constructor);  // "B"

TypeScript：基类中的多态返回类型？

1 个答案: