How to declare private abstract method in TypeScript?

Question

How to correctly define private abstract methods in TypeScript?

Here is a simple code:

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())
    }
}

This code does not work. How to fix it?

Answer 1

Quick aside, isCitrus should be this.isCitrus. On with the main show...

Abstract methods must be visible to sub-classes, because you are requiring the sub-class to implement the method.

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())
    }
}

If you want the method to be truly private, don't declare it on the base class.

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())
    }
}