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_;
}
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)
将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)
将r
设置为0.如果i
=,则为Ditto 2然后r
= 2,依此类推。