Question

只是想知道你是否可以在android中使用shift运算符我在尝试时遇到语法错误。运营商是＆gt;＆gt; ＆LT;＆LT; ＆GT;＆GT;＆GT; 。如果它不支持它是他们的Android sdk等价？

编辑：这是我正在使用的代码。我正在尝试进行每像素碰撞检测并尝试这样做。

public void getBitmapData(Bitmap bitmap1, Bitmap bitmap2){
     int[] bitmap1Pixels;
     int[] bitmap2Pixels;

    int bitmap1Height = bitmap1.getHeight();
    int bitmap1Width = bitmap1.getWidth();
    int bitmap2Height = bitmap1.getHeight();
    int bitmap2Width = bitmap1.getWidth();
    bitmap1Pixels = new int[bitmap1Height * bitmap1Width];
    bitmap2Pixels = new int[bitmap2Height * bitmap2Width];

    bitmap1.getPixels(bitmap1Pixels, 0, bitmap1Width, 1, 1, bitmap1Width - 1, bitmap1Height - 1); 
    bitmap2.getPixels(bitmap2Pixels, 0, bitmap2Width, 1, 1, bitmap2Width - 1, bitmap2Height - 1); 


 // Find the first line where the two sprites might overlap
    int linePlayer, lineEnemy;
    if (ninja.getY() <= enemy.getY()) {
        linePlayer = enemy.getY() - ninja.getY();
        lineEnemy = 0;
    } else {
        linePlayer = 0;
        lineEnemy = ninja.getY() - enemy.getY();
    }
    int line = Math.max(linePlayer, lineEnemy);

    // Get the shift between the two
    int x = ninja.getX() - enemy.getX();
    int maxLines = Math.max(bitmap1Height, bitmap2Height);
    for (; line <= maxLines; line ++) {
        // if width > 32, then you need a second loop here
        long playerMask = bitmap1Pixels[linePlayer];
        long enemyMask = bitmap2Pixels[lineEnemy];
        // Reproduce the shift between the two sprites
        if (x < 0) playerMask << (-x);
        else enemyMask << x;
        // If the two masks have common bits, binary AND will return != 0
        if ((playerMask & enemyMask) != 0) {
            // Contact!
            Log.d("pixel collsion","we have pixel on pixel");
        }

    }



}

Answer 1

如果你附加一个字符串，除非你把算术运算放在括号中，否则你会得到一个错误：


jcomeau@intrepid:/tmp$ cat test.java
public class test {
 public static void main(String args[]) {
  int test = 42;
  System.out.println("" + (test >> 1) + ", " + (test << 1) + ", " + (test >>> 1));
 }
}
jcomeau@intrepid:/tmp$ java test
21, 84, 21

Answer 2

Android使用的Java确实支持按位操作。 Here's a handy guide

android中的shift运算符语法错误

2 个答案: