Question

所以我有以下类声明：

abstract class A {
    constructor() {
        console.log("A.constructor() is called.");
        this.foo();
    }
    abstract foo();
}

abstract class B extends A {}

class C extends B {

    private bar = null;

    foo() {
        console.log("C.foo() is called.");
        this.bar = "bar";
    }

    getBar() {
        console.log("C.getBar() is called.");
        return this.bar;
    }

}

现在，我尝试通过以下操作对此进行调用：

let x = new C();
console.log(x.getBar());

以下是输出：

A.constructor() is called.
C.foo() is called.
C.getBar() is called.
null

我有完全相同的代码，但是在PHP中，它可以正常工作。有人可以解释一下我在想什么吗？

预先感谢，拉蒙

Answer 1

您将像上面的代码片段一样将Ts编译为JS，我记下了代码中的几点。

class A {
    constructor() {
        console.log("A.constructor() is called.");
        this.foo(); // 2. Set `this.bar = "bar"` => Ok
        // You can check bar value by add a `console.log(this.bar)` here
    }
}
class B extends A {
}
class C extends B {
    constructor() {
        super(...arguments); // 1. A constructor has been call at first
        this.bar = null; // 3. `bar` value has been `reset` to `null`
    }
    foo() {
        console.log("C.foo() is called.");
        this.bar = "bar";
    }
    getBar() {
        console.log("C.getBar() is called.");
        return this.bar;
    }
}
let x = new C();
console.log(x.getBar());

不能通过从继承的构造函数中调用本地方法来设置本地变量

1 个答案: