如何在TypeScript中声明私有的抽象方法？

Question

如何在TypeScript中正确定义私有抽象方法？

下面是一个简单的代码：

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()) 
    } 
} 

此代码不能正常工作。如何解决它？

Answer 1

快放一边，isCitrus应该是this.isCitrus。主显示...

抽象方法必须对于子类是可见的，因为您要求子类实现该方法。

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()) 
    } 
} 

如果您希望该方法是真正的私有方法，请不要在基类中声明该方法。

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()) 
    } 
}