如果語句似乎在跳到其他

Question

我正在嘗試編寫一個程序，該程序確定圓是否在內部/是否接觸矩形。 用戶為圓和半徑輸入中心點，為矩形輸入兩個對角點。

我不確定如何包含圓的所有點，以判斷矩形/矩形中至少有一個點。 有人知道該怎么做嗎？

當我運行當前程序時，我將有目的地輸入位於矩形內部的圓的點，並且應與我放置的if語句一起使用，但它會打印出錯誤的答案。

import java.util.Scanner;
public class lab4 {
public static void main(String[] args) {
    Scanner in = new Scanner(System.in);
    double cx, cy, x, y, r, p1x, p1y, p2x, p2y, max;//input 
    String a;

    System.out.print("Enter cx: ");
    cx = in.nextDouble();
    System.out.print("Enter cy: ");
    cy = in.nextDouble();    
    System.out.print("Enter r: ");
    r = in.nextDouble();

    System.out.println("Enter x value of point 1:");
    p1x = in.nextDouble();
    System.out.println("Enter y value of point 1:");
    p1y = in.nextDouble();
    System.out.println("Enter x value of point 2:");
    p2x = in.nextDouble();
    System.out.println("Enter y value of point 2:");
    p2y = in.nextDouble();


    max = p2x;
    if (p1x > max)
        max = p1x;

    max = p2y;
    if (p1y > max)
        max = p1y;

    if (cx >= p1x && cx <= p2x)
        a = "Circle is inside of Rectangle";
    if (cx >= p1x && cx <= p2x)
        a = "Circle is inside of Rectangle";
    if (cx+r >= p1x && cx+r <= p2x)
        a = "Circle is inside of Rectangle";
    if (cx-r >= p1x && cx-r <= p2x)
        a = "Circle is inside of Rectangle";
    if (cy >= p1y && cy <= p2y)
        a = "Circle is inside of Rectangle";
    if (cy >= p1y && cy <= p2y)
        a = "Circle is inside of Rectangle";
    if (cy+r >= p1y && cy+r <= p2y)
        a = "Circle is inside of Rectangle";
    if (cy-r >= p1y && cy-r <= p2y)
        a = "Circle is inside of Rectangle";
    else
        a = "Circle is outside of Rectangle";

    System.out.println(a); 

Answer 1

您的else語句僅以最后一個if語句為條件。 因此，如果最后一個if語句為false，則執行else語句。 您可能想要：

if ...
else if ...
else if ...
else

僅當所有先前的“ if”語句為false時，才執行else。

Answer 2

因為您沒有為每個條件使用else if ，所以最后一對if和else語句對將覆蓋之前的所有if 。

  
 
  
  
  
    if (cx >= p1x && cx <= p2x) a = "Circle is inside of Rectangle"; if (cx >= p1x && cx <= p2x) a = "Circle is inside of Rectangle"; if (cx+r >= p1x && cx+r <= p2x) a = "Circle is inside of Rectangle"; if (cx-r >= p1x && cx-r <= p2x) a = "Circle is inside of Rectangle"; if (cy >= p1y && cy <= p2y) a = "Circle is inside of Rectangle"; if (cy >= p1y && cy <= p2y) a = "Circle is inside of Rectangle"; if (cy+r >= p1y && cy+r <= p2y) a = "Circle is inside of Rectangle"; 
  
if (cy-r >= p1y && cy-r <= p2y)
    a = "Circle is inside of Rectangle";
else
    a = "Circle is outside of Rectangle";

對於每種選擇，請確保使用else if ，以確保僅執行其中一個塊。

if (cx >= p1x && cx <= p2x)
    a = "Circle is inside of Rectangle";
else if (cx >= p1x && cx <= p2x)
    a = "Circle is inside of Rectangle";
else if (cx+r >= p1x && cx+r <= p2x)
    a = "Circle is inside of Rectangle";
else if (cx-r >= p1x && cx-r <= p2x)
    a = "Circle is inside of Rectangle";
else if (cy >= p1y && cy <= p2y)
    a = "Circle is inside of Rectangle";
else if (cy >= p1y && cy <= p2y)
    a = "Circle is inside of Rectangle";
else if (cy+r >= p1y && cy+r <= p2y)
    a = "Circle is inside of Rectangle";
else if (cy-r >= p1y && cy-r <= p2y)
    a = "Circle is inside of Rectangle";
else
    a = "Circle is outside of Rectangle";

（此更正將解決當前問題，但是整個算法仍然不正確。）

Answer 3

正如其他人所說，您需要一系列if ... else if ... else if ... else才能使您的邏輯正常工作。

但是，有一種更簡單的方法。 要測試圓的任何部分是否接觸到矩形或在矩形內部，只需將矩形擴展圓的半徑，然后測試圓的中心是在擴展矩形的內部還是在擴展矩形上。 由於使用的是雙坐標，因此可以使用Rectangle.Double來完成所有繁重的工作：

public static void main(String[] args) {
    double cx, cy, r, p1x, p1y, p2x, p2y;

    // first input cx, cy, r, p1x, p1y, p2x, and p2y

    // construct a zero-width/height rectangle at p1
    Rectangle2D.Double p1 = new Rectangle2D.Double(p1x, p1y, 0, 0);

    // construct another one at p1
    Rectangle2D.Double p2 = new Rectangle2D.Double(p2x, p2y, 0, 0);

    // construct the union of the two
    Rectangle2D.Double rect = p1.createUnion(p2);

    // expand the rectangle
    rect.setBounds(rect.x - r, rect.y - r, rect.w + 2 * r, rect.h + 2 * r);

    // test for containment
    if (rect.contains(cx, cy) {
        a = "Circle is inside of Rectangle";
    } else {
        a = "Circle is outside of Rectangle";
    }
    System.out.println(a);
}

Answer 4

一些偽代碼：

將圓移到原點可簡化操作。 將矩形移位相同的數量。

p1x = p1x - cx
p2x = p2x - cx
p1y = p1y - cy
p2y - p2y - cy

x^2 + y^2 = r^2
y = +- sqrt( r^2 - x^2)

For x = -r to r

    y = + sqrt( r^2 - x^2 )

    if ( Inbounds(x,y) )return true;

    y = - sqrt(  r^2 - x^2 )

    if ( Inbounds(x,y) )return true;

End For

為了提高精度，您可以執行以下操作：

對於x = -r至r步驟0.01（使用雙精度並遞增0.01）

Answer 5

因為您將每個情況分開處理，而不使用else if ，所以if條件是如果if條件為true則覆蓋a的值，則else if與最后一個if語句有關，而不是全部。

我建議每個串聯結果變量a像這樣的，看看哪些條件是有效的：

if (cx >= p1x && cx <= p2x)
    a += "Circle is inside of Rectangle \n";
if (cx >= p1x && cx <= p2x)
    a += "Circle is inside of Rectangle\n";
if (cx+r >= p1x && cx+r <= p2x)
    a += "Circle is inside of Rectangle\n";
if (cx-r >= p1x && cx-r <= p2x)
    a += "Circle is inside of Rectangle\n";
if (cy >= p1y && cy <= p2y)
    a += "Circle is inside of Rectangle\n";
if (cy >= p1y && cy <= p2y)
    a += "Circle is inside of Rectangle\n";
if (cy+r >= p1y && cy+r <= p2y)
    a += "Circle is inside of Rectangle\n";
if (cy-r >= p1y && cy-r <= p2y)
    a += "Circle is inside of Rectangle\n";
else
    a += "Circle is outside of Rectangle\n";

或者，如果這不是您想要的，則將else if添加到所有這樣的if語句中：

if (cx >= p1x && cx <= p2x)
    a = "Circle is inside of Rectangle";
else if (cx >= p1x && cx <= p2x)
    a = "Circle is inside of Rectangle";
else if (cx+r >= p1x && cx+r <= p2x)
    a = "Circle is inside of Rectangle";
else if (cx-r >= p1x && cx-r <= p2x)
    a = "Circle is inside of Rectangle";
else if (cy >= p1y && cy <= p2y)
    a = "Circle is inside of Rectangle";
else if (cy >= p1y && cy <= p2y)
    a = "Circle is inside of Rectangle";
else if (cy+r >= p1y && cy+r <= p2y)
    a = "Circle is inside of Rectangle";
else if (cy-r >= p1y && cy-r <= p2y)
    a = "Circle is inside of Rectangle";
else
    a = "Circle is outside of Rectangle";