英文:
Calculating Area in Java; Order of Operations
问题
我正在尝试使用以下公式找到多边形的面积：
面积 = r^2 n sin( 2 π / n) / 2
其中 n 是边数，r 是半径。我认为我的代码产生的结果不正确。如果 n = 6，r = 4，我得到的面积是 24。我的代码如下：
import java.math.*;
public class RegularPolygon {
private int n; // 顶点数
private double r; // 多边形的半径
/**
* 这是完整的构造函数
* @param n 是顶点数
* @param r 是半径
*/
public RegularPolygon(int n, double r) {
super();
this.n = n;
this.r = r;
}
/**
* 默认构造函数。将边数和半径设置为零
*/
public RegularPolygon() {
super();
this.n = 0;
this.r = 0;
}
// 对顶点数 "n" 和半径 "r" 的获取和设置方法。
public int getN() {
return n;
}
public void setN(int n) {
this.n = n;
}
public double getR() {
return r;
}
public void setR(double r) {
this.r = r;
}
@Override
public String toString() {
return "RegularPolygon [n=" + n + ", r=" + r + "]";
}
public double area(){
float area;
// 该方法返回多边形的面积
// 根据需要使用 Math.PI 和 Math.sin
return area = (float) (Math.pow(r, 2)* n * ( Math.sin(Math.PI / n)/2));
}
我不清楚我的操作顺序出了什么问题。
英文:
I'm trying to find the area of a Polygon using the following formula:
Area = r^2 n sin( 2 π / n) / 2
where n is the number of sides and r is the radius. I do not think my code is producing the correct result. If n= 6 and r = 4, i'm getting an area of 24. My code is as follows:
import java.math.*;
public class RegularPolygon {
private int n; // number of vertices
private double r; //radius of the polygon
/**
* This is the full constructor
* @param n is the number of vertices
* @param r is the radius
*/
public RegularPolygon(int n, double r) {
super();
this.n = n;
this.r = r;
}
/**
* default constructor. Sets number of sides and radius to zero
*/
public RegularPolygon() {
super();
this.n = 0;
this.r = 0;
}
//getters and setters for Number of vertices "n" and radius "r".
public int getN() {
return n;
}
public void setN(int n) {
this.n = n;
}
public double getR() {
return r;
}
public void setR(double r) {
this.r = r;
}
@Override
public String toString() {
return "RegularPolygon [n=" + n + ", r=" + r + "]";
}
public double area(){
float area;
//this method returns the area of the polygon
//Use Math.PI and Math.sin as needed
return area = (float) (Math.pow(r, 2)* n * ( Math.sin(Math.PI / n)/2));
It is unclear to me where my order of operations is messed up.
答案1
得分: 0
你没有正确翻译这个公式。你需要使用 return Math.pow(r, 2) * n * Math.sin(2 * Math.PI / n) / 2; 正如评论中所指出的，你漏掉了一个 2 并写成了 Math.sin(Math.PI / n)。这与操作顺序无关，因为这只涉及乘法和除法。
英文:
You're not translating the formula correctly. You need return Math.pow(r, 2) * n * Math.sin(2 * Math.PI / n) / 2; As forpas pointed out in the comments, you're missing a 2 and saying Math.sin(Math.PI / n) This has nothing to do with order of operations, since it's all just multiplication and division.
答案2
得分: 0
你不应该将类型转换为浮点型，因为你已经声明了方法的返回类型为双精度型。
return Math.pow(r, 2) * n * Math.sin(2 * Math.PI/n) / 2;
我认为上述代码将满足你的需求。
英文:
You should not type cast to float as you have declared the return type of the method as double
return Math.pow(r,2) * n * Math.sin(2 * Math.PI/n) / 2;
I think the above code will satisfy your need.
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