子类的构造函数中的默认值传递给超类？

Question

我有一个带有构造函数的类形状，在其构造函数中采用 (int Sides, Coodinates c, Color co)，在类正方形中我不想在构造函数中有“int Sides”，我该怎么做这个？

//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 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，你是对的。但是，您还需要分配 w 和 l 的值。