如何在TypeScript中聲明私有抽象方法？

Question

如何在TypeScript中正確定義私有抽象方法？

這是一個簡單的代碼：

abstract class Fruit {
    name: string;
    constructor (name: string) {
        this.name = name
    }
    abstract private hiFrase (): string;
}

class Apple extends Fruit {
    isCitrus: boolean;
    constructor(name: string, isCitrus: boolean) {
        super(name);
        this.isCitrus = isCitrus;
    }

    private hiFrase(): string {
        return "Hi! I\'m an aplle and my name is " + this.name + " and I'm " + (isCitrus ? "" : " not ") + "citrus";
    }

    public sayHi() {
        alert(this.hiFrase())
    }
}

此代碼不起作用。 怎么解決？

Answer 1

快一下， isCitrus應該是this.isCitrus 。 隨着主要節目......

抽象方法必須對子類可見，因為您需要子類來實現該方法。

abstract class Fruit {
    name: string;
    constructor (name: string) {
        this.name = name
    }
    protected abstract hiFrase(): string;
}

class Apple extends Fruit {
    isCitrus: boolean;
    constructor(name: string, isCitrus: boolean) {
        super(name);
        this.isCitrus = isCitrus;
    }

    protected hiFrase(): string {
        return "Hi! I\'m an aplle and my name is " + this.name + " and I'm " + (this.isCitrus ? "" : " not ") + "citrus";
    }

    public sayHi() {
        alert(this.hiFrase())
    }
}

如果您希望該方法是真正的私有，請不要在基類上聲明它。

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

class Apple extends Fruit {
    isCitrus: boolean;
    constructor(name: string, isCitrus: boolean) {
        super(name);
        this.isCitrus = isCitrus;
    }

    private hiFrase(): string {
        return "Hi! I\'m an aplle and my name is " + this.name + " and I'm " + (this.isCitrus ? "" : " not ") + "citrus";
    }

    public sayHi() {
        alert(this.hiFrase())
    }
}