java-如何设计从Rectangle类继承的Square类
作者:互联网
因此,我正在尝试为我的其中一篇教程编写一些代码.输入和预期输出是这样的:
> 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如何工作.如何修改代码,使输入返回正确的输出?我认为错误仅存在于Square类中,否则代码将编译.
提前致谢.
解决方法:
在设计方面,我认为构造函数Rectance(double width)对于矩形而言是不自然的,因此将其删除.
Square的构造函数应如下所示:
public Square(double side) {
super(side,side); // width == height
this.shapeName = "Square";
}
为了进一步阐明继承,您还可以替换行this.shapeName =“ Rectangle”;与this.shapeName = getClass().getSimpleName();并删除this.shapeName =“ Square”;来自Square的构造函数.
标签:inheritance,abstract-class,java 来源: https://codeday.me/bug/20191025/1925741.html