[英]Understanding java code (abstract classes, extends function)
Im trying to understand this code ... 我试图理解这段代码......
Here is the original code: 这是原始代码:
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;
}
}
Can someone please help me understand how this code returns 28. But also explain the part where I am getting stuck... 有人可以帮我理解这段代码如何返回28.但也解释了我被卡住的部分......
When dry running, I get stuck here: 干运行时,我卡在这里:
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_;
}
The main
loop produces, in order, Circle(0)
, Polygon(1,1)
, Circle(2)
, Polygon(3,3)
, Circle(4)
; main
循环依次产生Circle(0)
, Polygon(1,1)
, Circle(2)
, Polygon(3,3)
, Circle(4)
; this is because i % 2
is 0 for even numbers and 1 for odd numbers. 这是因为对于偶数, i % 2
为0,对于奇数则为1。 For the polygons, measure
returns 2*(x+y), so for (1,1) you have 2*(1+1) = 4 and for (3,3) you have 2 * (3+3) = 12. For the circles, measure
returns 2*r, so for (0) you have 2 * 0 = 0, for (2) you have 2 * 2 = 4, and for (4) you have 2 * 4 = 8. 0+4+4+12+8 = 28. 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 + 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。 + 4 + 4 + 12 + 8 = 28。
For the spot where you get lost, getShape(i)
passes the value of i
to getShape
; 对于您迷路现场, getShape(i)
传递的价值i
来getShape
; when i
= 0 this creates a Circle, so 0
is passed to the Circle constructor, and then Circle(0)
sets r
to 0. Ditto if i
= 2 then r
= 2, and so on. 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
= 2则ditto,则r
= 2,依此类推。
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