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此步骤必须返回 Shape 类型的结果

发布时间: 2011-11-30 21:16:57 作者: rapoo

此方法必须返回 Shape 类型的结果?
import java.util.Random;
class Shape{
void draw(){}
void erase(){}
}

class Circle extends Shape{
void draw(){
System.out.println( "Circle.draw() ");
}
void erase(){
System.out.println( "Circle.erase ");
}
}

class Square extends Shape{
void draw(){
System.out.println( "Square.draw() ");
}
void erase(){
System.out.println( "Square.erase ");
}
}

class Triangle extends Shape{
void draw(){
System.out.println( "Triangle.draw() ");
}
void erase(){
System.out.println( "Triangle.erase ");
}
}

class RandomShapeGenerator{
private Random random = new Random();
public Shape next(){//error,此方法必须返回 Shape 类型的结果
switch(random.nextInt(3)){
//default:
case 0:return new Circle();
case 1:return new Square();
case 2:return new Triangle();
}
}
}
public class Duotai{
private static RandomShapeGenerator gen = new RandomShapeGenerator();
public static void main(String[] args){
Shape[] s = new Shape[4];
for( int i = 0; i < s.length; i++ ){
s[i] = gen.next();
for( int j = 0; j < s.length; j++ ){
s[j].draw();
}
}
}
}

[解决办法]
实际返回的类型都是Shape 的子类


[解决办法]
error?
[解决办法]
public Shape next(){//error,此方法必须返回 Shape 类型的结果
switch(random.nextInt(3)){
case 0:return new Circle();
case 1:return new Square();
case 2:return new Triangle();
default: return new Shape();
}

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