Java代碼爲三角形區域返回0

Question

我正在介紹Java課程，我的任務是編寫一個程序，該程序需要在三角形的三邊並返回類型以及區域。雙方必須根據作業單是整數，所以它被證明是一個痛苦的測試它，並提出整數比率來匹配每種類型。

無論如何，我遇到的真正問題是它有時會給我正確的區域，有時它不會。例如6/8/10返回面積爲0.

任何想法/一般提示，以改善源？到目前爲止，我看起來像意大利麪條，但我有點受限，因爲我只允許使用我們到目前爲止所講的內容。

package trianglesides; 
import javax.swing.*; 
/* 
This program will receive three numbers from a user describing the lengths of 
the sides of a triangle. It will return and display the type of triangle and 
the area. 
*/ 
public class triangle_Sides { 
    public static void main(String[] args) { 
     int side1 = Integer.parseInt(JOptionPane.showInputDialog 
      ("Enter the integer length of side 1")); 
     int side2 = Integer.parseInt(JOptionPane.showInputDialog 
      ("Enter the integer length of side 2")); 
     int side3 = Integer.parseInt(JOptionPane.showInputDialog 
      ("Enter the integer length of side 3")); 
     int s = (side1 + side2 + side3)/2; 
     double area = Math.sqrt(s*(s-side1)*(s-side2)*(s-side3)); 
     if (side1 >= (side2 + side3) || 
      side2 >= (side1 + side3) || 
      side3 >= (side1 + side2)) { 
      JOptionPane.showMessageDialog(null, "That's no triangle."); 
     } 
     else { 
/*Equilateral*/ 
     if ((side1 == side2) && (side2 == side3)) { 
      JOptionPane.showMessageDialog(null, "This is an equilateral triangle with an area of " + area); 
     } 
/*Right and isosceles*/ 
     else if ((side1*side1) == ((side2*side2) + (side3*side3)) || 
       (side2*side2) == ((side1*side1) + (side3*side3)) || 
       (side3*side3) == ((side2*side2) + (side1*side1))) { 
      if (side1 == side2 || 
       side2 == side3 || 
       side3 == side1){ 
       JOptionPane.showMessageDialog(null, "This is right and isosceles triangle with an area of " + area); 
      } 
      else { 
       JOptionPane.showMessageDialog(null, "This is a right triangle with an area of " + area); 
      } 
     } 
/*Obtuse and isosceles*/ 
     else if ((side1*side1) > ((side2*side2) + (side3*side3)) || 
       (side2*side2) > ((side1*side1) + (side3*side3)) || 
       (side3*side3) > ((side2*side2) + (side1*side1))) { 
      if (side1 == side2 || 
       side2 == side3 || 
       side3 == side1){ 
       JOptionPane.showMessageDialog(null, "This is obtuse and isosceles triangle with an area of " + area); 
      } 
      else { 
       JOptionPane.showMessageDialog(null, "This is an obtuse triangle with an area of " + area); 
      } 
     } 
/*Acute and isosceles*/ 
     else if ((side1*side1) < ((side2*side2) + (side3*side3)) || 
       (side2*side2) < ((side1*side1) + (side3*side3)) || 
       (side3*side3) < ((side2*side2) + (side1*side1))) { 
      if (side1 == side2 || 
       side2 == side3 || 
       side3 == side1){ 
       JOptionPane.showMessageDialog(null, "This is acute and isosceles triangle with an area of " + area); 
      } 
      else { 
       JOptionPane.showMessageDialog(null, "This is an acute triangle with an area of " + area); 
      } 
     } 
     } 
    } 
} 

Answer 1

發生這種情況，因爲變量小號不能是整數類型

變化

int s = (side1 + side2 + side3)/2; 

到

double s = (side1 + side2 + side3)/2.0; 

Answer 2

int s = (side1 + side2 + side3)/2; 

整數的舍入將ロ並減半。這不是你想要的。除以2.0並將結果代入double。

Math.sqrt(s*(s-side1)*(s-side2)*(s-side3)) 

你正在做整數運算這個產品，它將環繞而不是失去精度。這不是你想要的。此外，這是famously not the expression of Heron's formula you want on computers.

對於一堆比較，你可以做得更好，首先按順序排序邊。除此之外，它對我來說看起來很合理。您可能會考慮將「讀取輸入和輸出」部分與「執行計算」部分分開，以便您可以更輕鬆地調試和測試計算。