如何设计一个继承自 Rectangle 类的 Square 类

Question

所以我正在尝试为我的一个教程编写一些代码。输入和预期输出如下：

> Square s = new Square(5);

> s.toString();

< Square with area 25.00 and perimeter 20.00

以下是我的代码：

abstract class Shape {

    protected String shapeName;

    public abstract double getArea();

    public abstract double getPerimeter();

    @Override

    public String toString() {

        return shapeName + " with area " + String.format("%.2f", getArea()) +

            " and perimeter " + String.format("%.2f", getPerimeter());

    }

}

class Rectangle extends Shape {

    protected double width;

    protected double height;

    public Rectangle(double width) {

        this.shapeName = "Rectangle";

        this.width = this.height = width;

    }

    public Rectangle(double width, double height) {

        this.shapeName = "Rectangle";

        this.width = width;

        this.height = height;

    }

    public double getArea() {

        return width * height;

    }

    public double getPerimeter() {

        return 2 * (width + height);

    }

}

class Square extends Rectangle {

    public Square(double side) {

        this.shapeName = "Square";

        this.width = this.height = side;

    }

}

问题是当我尝试编译它时，会发生此错误：

error: no suitable constructor found for Rectangle(no arguments)

    public Square(double side) {

                               ^

    constructor Rectangle.Rectangle(double) is not applicable

      (actual and formal argument lists differ in length)

    constructor Rectangle.Rectangle(double,double) is not applicable

      (actual and formal argument lists differ in length)

我不确定在这种情况下继承是如何工作的。我如何修改我的代码以使输入返回正确的输出？我认为错误仅存在于 Square 类中，因为代码以其他方式编译。

Answer 1

在设计方面，我觉得构造函数 Rectance( double width ) 对于矩形来说是不自然的，会删除它。Square 的构造函数应如下所示：&nbsp;public Square(double side) {&nbsp; &nbsp; &nbsp; &nbsp; super(side,side);&nbsp; &nbsp;// width == height&nbsp; &nbsp; &nbsp; &nbsp; this.shapeName = "Square";&nbsp;}为了进一步阐明继承，您还可以this.shapeName= "Rectangle";使用from的构造函数替换this.shapeName= getClass().getSimpleName();并删除该行。this.shapeName = "Square";Square

Answer 2

您必须显式调用super继承类中的构造函数。如果你想考虑@runec 的回答（是的，我赞成它，因为它很好），你可能想删除只有一个参数的构造函数，Rectangle并使Square句柄成为一个有 4 个等边的矩形。public class Rectangle extends Shape {&nbsp; &nbsp; protected double width;&nbsp; &nbsp; protected double height;&nbsp; &nbsp; public Rectangle(double width, double height) {&nbsp; &nbsp; &nbsp; &nbsp; this.shapeName = "Rectangle";&nbsp; &nbsp; &nbsp; &nbsp; this.width = width;&nbsp; &nbsp; &nbsp; &nbsp; this.height = height;&nbsp; &nbsp; }&nbsp; &nbsp; public double getArea() {&nbsp; &nbsp; &nbsp; &nbsp; return width * height;&nbsp; &nbsp; }&nbsp; &nbsp; public double getPerimeter() {&nbsp; &nbsp; &nbsp; &nbsp; return 2 * (width + height);&nbsp; &nbsp; }}这里你有Square只需要一个double参数的 ：public class Square extends Rectangle {&nbsp; &nbsp; public Square(double side) {&nbsp; &nbsp; &nbsp; &nbsp; super(side, side);&nbsp; &nbsp; &nbsp; &nbsp; this.shapeName = "Square";&nbsp; &nbsp; &nbsp; &nbsp; this.width = this.height = side;&nbsp; &nbsp; }}这个答案只是为了展示完整的代码，完整的想法来自@runec。