有哪位大神帮写一个，蟹蟹

[南浔[思埠]]

shape接口中有一个抽象方法area(),方法可以接收一个double类型的参数,返回一个double 类型的结果

[能能火柴人]

public interface shape { public double area(double a); }

[南浔[思埠]]

引用 1 楼 能能火柴人的回复:

public interface shape { public double area(double a); }

设计一个shape接口和它的两个实现类square和circle类，要求:(刚才那个是第一个要求)第二个要求:square和ciecle中实现了shape接口的area()方法，分别求正方形和圆形的面积并返回。求大神帮写一个

[能能火柴人]

public class square implements shape{ @Override public double area(double a) { return a*a; } } public class ciecle implements shape{ @Override public double area(double a) { return 3.14*a*a; } }

[倾听心跳的频率灬]

public interfacpublic class Square implements Shape{ public double area(double x) { return x*x; } } public class Square implements Shape{ public double area(double x) { return x*x; } } public class Circle implements Shape{ public double area(double x) { return Math.PI*(x/2)*(x/2); } } public class Test { public static void main(String[] args) { Square s=new Square(); double s0=s.area(5); System.out.println("正方形的面积是："+s0); Circle c=new Circle(); double c0=c.area(10); System.out.println("圆的面积为："+c0); } }

[倾听心跳的频率灬]

public interface Shape { double area(double x); } public class Square implements Shape{ public double area(double x) { return x*x; } } public class Circle implements Shape{ public double area(double x) { return Math.PI*(x/2)*(x/2); } } public class Test { public static void main(String[] args) { Square s=new Square(); double s0=s.area(5); System.out.println("正方形的面积是："+s0); Circle c=new Circle(); double c0=c.area(10); System.out.println("圆的面积为："+c0); } }

[倾听心跳的频率灬]

第一次回复的那个复制错了没注意

[低级攻城狮]

哈哈，你们可真闲

[菜鸟程序员chen]

哈哈，无聊过来刷一刷，好多问题，像是看到了以前的自己，，还有小伙子，加油啊