理解java代码(抽象类,扩展函数)

时间:2013-04-17 17:09:25

标签: java

我试图理解这段代码......

以下是原始代码:

public class understanding {
    public static void main(String[] args) {
        int n = 5;
        Shape[] s = new Shape[n];
        for (int i = 0; i < n; i++) {
            s[i] = getShape(i);
            System.out.println(s[i]);
        }
        System.out.println(compute(s));
    }

    public static Shape getShape(int i) {
        if (i % 2 == 0) {
            return new Circle(i);
        } else {
            return new Polygon(i, i);
        }
    }

    public static double compute(Shape[] a) {
        double nc = 0;
        for (int i = 0; i < a.length; i++) {
            nc += a[i].getMeasure();
            System.out.println(a[i].getMeasure());
        }
        return nc;
    }
}

public abstract class Shape {
    protected abstract double getMeasure();
}

public class Polygon extends Shape {
    private double x;
    private double y;

    Polygon(double x_, double y_) {
        x = x_;
        y = y_;
    }

    public double getMeasure() {
        return 2 * (x + y);
    }
}

public class Circle extends Shape {
    private double r;

    Circle(double r_) {
        r = r_;
    }

    public double getMeasure() {
        return 2 * r;
    }
}

有人可以帮助我理解这段代码如何返回28.但也解释了我被卡住的部分......

干涸的时候,我被困在这里:

for(int i=0;i<n;i++){  // i=0, i<5, i++
    s[i] = getShape[i] // i=0, go to getShape class
...

getShape(i) //i=0
if (i%2==0) //TRUE so..
    return new Circle(i); //bc the boolean is True, go to Circle() class
...

Circle extends Shape{
    private double r; 
    Circle(double r_){ //I get lost here.. is r=0 now (because r_=i and r=r_)?
        r = r_;
    }

1 个答案:

答案 0 :(得分:2)

main循环按顺序生成Circle(0)Polygon(1,1)Circle(2)Polygon(3,3)Circle(4);这是因为i % 2对于偶数是0而对于奇数是1。对于多边形,measure返回2 *(x + y),因此对于(1,1),您有2 *(1 + 1)= 4,对于(3,3),您有2 *(3 + y) 3)= 12.对于圆圈,measure返回2 * r,所以对于(0)你有2 * 0 = 0,对于(2)你有2 * 2 = 4,对于(4)你有2 * 4 = 8. 0 + 4 + 4 + 12 + 8 = 28。

对于迷路的地方,getShape(i)i的值传递给getShape;当i = 0时,这会创建一个Circle,因此将0传递给Circle构造函数,然后Circle(0)r设置为0.如果i =,则为Ditto 2然后r = 2,依此类推。