本文介绍了不能让Rect.intersects(REC,REC2)工作的处理方法,对大家解决问题具有一定的参考价值,需要的朋友们下面随着小编来一起学习吧!
问题描述
出于某种原因,我的程序将无法通过获得Rect.intersects(RE,hitcore)
。如果我使用的System.out.println()re.flattenToString()
,我看到两个矩形相交,但它不会返回true
。
我也使用 re.intersect(hitcore)
,但仍没有尝试过。帮助?
进口android.graphics.Rect;市民阶级敌人{
公共矩形重,hitCore;私人布尔命中= FALSE;
公敌() {
重=新的Rect(0,0,0,0);
hitCore =新的Rect(0,0,0,0);}公共无效更新(){ re.set(+的centerX leftX,centerY + TOPY,rightX,BOTY); //这些都是由其他类设置
hitCore.set(ship.getCenterX()+ 3,ship.getCenterY()+ 10,93,20); //System.out.println(re.flattenToString()+Rect1的);
//System.out.println(hitCore.flattenToString()+RECT2); checkHit(hitCore);}私人无效checkHit(矩形hitCore){
如果(Rect.intersects(hitCore,RE)){
命中= TRUE;
}
}
}
解决方案
我想通了。我用x,y和宽度,高度设定矩形。实际参数是左上角坐标和右下角坐标。
rect.set(X1,Y1,X2,Y2);
For some reason my program won't get through Rect.intersects(re,hitcore)
. If I use System.out.println()re.flattenToString()
, I see that the two rectangles intersect but it won't return true
.
I also tried using re.intersect(hitcore)
, but still nothing. Help?
import android.graphics.Rect;
public class Enemy {
public Rect re, hitCore;
private boolean hit = false;
public Enemy() {
re = new Rect(0, 0, 0, 0);
hitCore = new Rect(0, 0, 0, 0);
}
public void update() {
re.set(centerX + leftX, centerY + topY, rightX, botY); // these are set by another class
hitCore.set(ship.getCenterX() + 3, ship.getCenterY() + 10, 93, 20);
//System.out.println(re.flattenToString() + " rect1");
//System.out.println(hitCore.flattenToString() + " rect2");
checkHit(hitCore);
}
private void checkHit(Rect hitCore) {
if (Rect.intersects(hitCore, re)) {
hit = true;
}
}
}
解决方案
I figured it out. I was setting up rectangle using x,y and width,height. The actual parameters are the top left corner coordinate and bottom right corner coordinate.
rect.set(x1,y1,x2,y2);
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