实施例

Question

我正在研究一个程序，将三角形分类为三角形的各个点（等边，斜角，右边等）。我已经完成了99％，但有一个Else-if声明无法正常工作。

对于最终的测试运行，我们应该输入点（0,0）（0,5）（5,0），然后得到输出＆＃34;等边，右边＆＃34;。但是，我的代码似乎只是跳过该语句并继续执行下一个if语句，该语句将三角形分类为scalene。

提前致谢。我是一个新手编码器，所以它可能比较简单，可以躲避我的眼睛。

这是整个代码。

public class TriangleCalculator {

    @SuppressWarnings("empty-statement")
    public static void main(String[] args) 
    {
      System.out.println("Enter the x- and y- coordinates of the first point");
      Scanner keyBoard = new Scanner (System.in);
        double coordOneX = keyBoard.nextDouble();
        double coordOneY = keyBoard.nextDouble();
      System.out.println("Enter the x- and y- coordinates of the second point");
        double coordTwoX = keyBoard.nextDouble();
        double coordTwoY = keyBoard.nextDouble();
      System.out.println("Enter the x- and y- coordinates of the third point");
        double coordThreeX = keyBoard.nextDouble();
        double coordThreeY = keyBoard.nextDouble(); 

        if((coordOneY - coordTwoY) * (coordTwoX - coordThreeX) == (coordTwoY - coordThreeY) * (coordOneX - coordTwoX))
        {
           System.out.printf ("D1 ={(%.3f,%.3f) , D2 = (%.3f,%.3f) , D3 = (%.3f,%.3f) are colinear and cannot be vertices of a triangle.}%n",
                coordOneX, coordOneY, coordTwoX, coordTwoY, coordThreeX, coordThreeY);
        }
        else if((coordOneY - coordTwoY) * (coordTwoX - coordThreeX) != (coordTwoY - coordThreeY) * (coordOneX - coordTwoX))
        {    
            System.out.printf ("Triangle ={(%.3f,%.3f) , (%.3f,%.3f) , (%.3f,%.3f) }%n",
                coordOneX, coordOneY, coordTwoX, coordTwoY, coordThreeX, coordThreeY);

            double distanceOne = Math.sqrt(Math.pow(coordOneX - coordTwoX,2) + Math.pow(coordOneY - coordTwoY,2));
                System.out.printf("Distance (P1,P2) = %.3f%n", distanceOne);
            double distanceTwo = Math.sqrt(Math.pow(coordTwoX - coordThreeX,2) + Math.pow(coordTwoY - coordThreeY,2));
                System.out.printf("Distance (P2,P3)= %.3f%n", distanceTwo);
            double distanceThree = Math.sqrt(Math.pow(coordOneX - coordThreeX,2) + Math.pow(coordOneY - coordThreeY,2));
                System.out.printf("Distance (P1,P3)= %.3f%n", distanceThree);   

            double perimeter = distanceOne + distanceTwo + distanceThree;
                System.out.printf("Perimeter = %.3f %n", perimeter);

            double s = perimeter / 2;
                double area = Math.sqrt(s * (s - distanceOne) * (s - distanceTwo) *(s - distanceThree));
                System.out.printf("Area = %.3f %n", area);

            double side1 = distanceOne;
            double side2 = distanceTwo;
            double side3 = distanceThree;               
            String classification = ("");

            if(side1 == side2)
                classification = "isosceles";
            else if(side2==side3)
                    classification = "equilateral";
            else if (Math.abs(side1*side1 + side2*side2 - side3*side3) < 1E-9)
                classification = "isoceles, right";
            if(side1 == side3||side2 == side3)
                 classification = "isosceles";
            else classification = "scalene";
            if(side1*side1 + side2*side2 == side3*side3)
                classification = "scalene, right";


        System.out.println("Clasification(s): " +classification);

这是有问题的代码的特定部分：

        if(side1 == side2)
            classification = "isosceles";
        else if(side2==side3)
                classification = "equilateral";
        else if (Math.abs(side1*side1 + side2*side2 - side3*side3) < 1E-9)
            classification = "isoceles, right";
        if(side1 == side3||side2 == side3)
             classification = "isosceles";
        else classification = "scalene";
        if(side1*side1 + side2*side2 == side3*side3)
            classification = "scalene, right";

Answer 1

由于双精度数和浮点数是浮点数，所以不应该将双精度数与==进行比较，这是因为你不能真正100％确定或保证你赋予float或double的数字是完全正确的在内存中映射的内容......

在您的情况下，请使用Double.compare并查看此article

实施例

// compares the two specified double values
double d1 = 15.45;
double d2 = 11.50;
int retval = Double.compare(d1, d2);

根据文件：

JLS 15.21.1. Numerical Equality Operators == and !=：

根据IEEE 754标准规范确定的浮点比较结果如下：

根据IEEE 754标准的规则执行浮点相等测试：



如果任一操作数为NaN，则==的结果为false，但！=的结果为true。实际上，当且仅当x的值是NaN时，测试x！= x才为真。方法Float.isNaN和Double.isNaN也可用于测试值是否为NaN。



正零和负零被认为是相等的。例如，-0.0 == 0.0为真。



否则，等于运算符会将两个不同的浮点值视为不相等。特别是，有一个值代表正无穷大，一个值代表负无穷大;每个比较仅与自身相等，每个比较不等于所有其他值。

Else-If语句输出不正确

1 个答案:

实施例

根据文件：