기본 클래스의 멤버에 액세스

기본 클래스의 멤버에 액세스

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TypeScript 사이트의 놀이터에서 상속 예제를 참조하십시오.

class Animal {
    public name;
    constructor(name) { 
        this.name = name;
    }
    move(meters) {
        alert(this.name + " moved " + meters + "m.");
    }
}

class Snake extends Animal {
    constructor(name) { super(name); }
    move() {
        alert("Slithering...");
        super.move(5);
    }
}

class Horse extends Animal {
    constructor( name) { super(name); }
    move() {
        alert(super.name + " is Galloping...");
        super.move(45);
    }
}

var sam = new Snake("Sammy the Python")
var tom: Animal = new Horse("Tommy the Palomino")

sam.move()
tom.move(34)

한 줄의 코드를 변경했습니다 : Horse.move(). 거기에 액세스 super.name하고 싶지만 그 결과가 반환 undefined됩니다. IntelliSense에서 사용할 수 있고 TypeScript가 제대로 컴파일되지만 작동하지 않는다고 제안합니다.

어떤 아이디어?

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작업 예. 아래 참고.

class Animal {
    constructor(public name) {
    }

    move(meters) {
        alert(this.name + " moved " + meters + "m.");
    }
}

class Snake extends Animal {
    move() {
        alert(this.name + " is Slithering...");
        super.move(5);
    }
}

class Horse extends Animal {
    move() {
        alert(this.name + " is Galloping...");
        super.move(45);
    }
}

var sam = new Snake("Sammy the Python");
var tom: Animal = new Horse("Tommy the Palomino");

sam.move();
tom.move(34);

1. 공용 변수에 이름을 수동으로 할당 할 필요가 없습니다. public name생성자 정의에서 사용하면 이 작업이 수행됩니다.

2. super(name)전문 클래스에서 전화 할 필요가 없습니다 .

3. this.name작품 사용 .

사용시주의 사항 super.

이것은 언어 사양의 섹션 4.9.2 에서 자세히 다룹니다 .

The behaviour of the classes inheriting from Animal is not dissimilar to the behaviour in other languages. You need to specify the super keyword in order to avoid confusion between a specialised function and the base class function. For example, if you called move() or this.move() you would be dealing with the specialised Snake or Horse function, so using super.move() explicitly calls the base class function.

There is no confusion of properties, as they are the properties of the instance. There is no difference between super.name and this.name - there is simply this.name. Otherwise you could create a Horse that had different names depending on whether you were in the specialized class or the base class.

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You are incorrectly using the super and this keyword. Here is an example of how they work:

class Animal {
    public name: string;
    constructor(name: string) { 
        this.name = name;
    }
    move(meters: number) {
        console.log(this.name + " moved " + meters + "m.");
    }
}

class Horse extends Animal {
    move() {
        console.log(super.name + " is Galloping...");
        console.log(this.name + " is Galloping...");
        super.move(45);
    }
}

var tom: Animal = new Horse("Tommy the Palomino");

Animal.prototype.name = 'horseee'; 

tom.move(34);
// Outputs:

// horseee is Galloping...
// Tommy the Palomino is Galloping...
// Tommy the Palomino moved 45m.

Explanation:

1. The first log outputs super.name, this refers to the prototype chain of the object tom, not the object tom self. Because we have added a name property on the Animal.prototype, horseee will be outputted.
2. The second log outputs this.name, the this keyword refers to the the tom object itself.
3. The third log is logged using the move method of the Animal base class. This method is called from Horse class move method with the syntax super.move(45);. Using the super keyword in this context will look for a move method on the prototype chain which is found on the Animal prototype.

Remember TS still uses prototypes under the hood and the class and extends keywords are just syntactic sugar over prototypical inheritance.

참고URL : https://stackoverflow.com/questions/13121431/accessing-member-of-base-class