传递给超类的子类构造函数中的默认值？

Question

我有一个带有构造函数的类形状，它在构造函数中采用（int Sides，Coodinates c，Color co），在类方块中我不想在构造函数中使用“int Sides”，我该怎么做？

//Shape
public Shape(int sides, Coordinates c, Color co) {
    this.sides = sides;
    this.c = c;
    this.co = co;
}

//Square
//Like this is gives me an error saying I need to include sides into my
//constructor. I don't want to request the number of sides when creating
//a square, it will always be 4, how do I do this?
public Square(Coordinates c, Color co, int w, int l) {
    //blah blah
}

假设我必须使用super（4，c，co）传递值4，我是否正确？

Answer 1

public Square(Coordinates c, Color co, int w, int l) {
    super(4, c, co);
    this.w = w;
    this.l = l;
}

是的，你会是对的。 但是，您还需要指定w和l的值。