如题↑↑↑
- Java code
package eclipsecookbook;import org.eclipse.swt.SWT;import org.eclipse.swt.graphics.Region;import org.eclipse.swt.widgets.Button;import org.eclipse.swt.widgets.Display;import org.eclipse.swt.widgets.Event;import org.eclipse.swt.widgets.Listener;import org.eclipse.swt.widgets.Shell;public class NonRectanglegularClass { public static void main(String[] args) { final Display display = new Display(); final Shell shell = new Shell(display, SWT.NO_TRIM); Region region = new Region(); region.add(createCircle(50, 50, 50)); region.subtract(createCircle(50, 50, 20)); shell.setRegion(region); shell.setSize(region.getBounds().width, region.getBounds().height); shell.setBackground(display.getSystemColor(SWT.COLOR_BLUE)); Button button = new Button(shell, SWT.PUSH); button.setText("Exit"); button.setBounds(35, 6, 35, 20); button.addListener(SWT.Selection, new Listener() { @Override public void handleEvent(Event arg0) { shell.close(); } }); shell.open(); while (!shell.isDisposed()) { if (!display.readAndDispatch()) { display.sleep(); } } display.dispose(); } static int[] createCircle(int xOffset, int yOffset, int radius) { int[] circlePoints = new int[10 * radius]; for (int loopIndex = 0; loopIndex < 2* radius + 1; loopIndex++) { int xCurrent = loopIndex - radius; int yCurrent = (int) Math.sqrt(radius * radius - xCurrent * xCurrent); int doubleLoopIndex = 2 * loopIndex; circlePoints[doubleLoopIndex] = xCurrent + xOffset; circlePoints[doubleLoopIndex + 1] = yCurrent + yOffset; circlePoints[10 * radius - doubleLoopIndex - 2] = xCurrent + xOffset; circlePoints[10 * radius - doubleLoopIndex - 1] = -yCurrent + yOffset; } return circlePoints; }}------解决方案--------------------------------------------------------
你已经返回一个数组点阵了，循环取出来就行了啊
------解决方案--------------------------------------------------------
楼主你想问的是，程序是如何计算出点阵的么？
是用直角三角计算公式来算的：
for (int loopIndex = 0; loopIndex < 2* radius + 1; loopIndex++) { // 注意它循环到半径×2
int xCurrent = loopIndex - radius; // 得到x坐标，结合循环起至条件，相当于从 (圆心-半径) --> (圆心+半径)
int yCurrent = int (int) Math.sqrt(radius * radius - xCurrent * xCurrent); // 直角三角形，已知斜边和直角边的边长，求另一个直角边
这种计算过程，每次实际上一个x坐标会计算出两个y坐标（以圆心所在Y轴的轴对称上），所以后续赋值的时候，是两个点：
circlePoints[doubleLoopIndex] = xCurrent + xOffset;
circlePoints[doubleLoopIndex + 1] = yCurrent + yOffset; // 这个是正的
circlePoints[10 * radius - doubleLoopIndex - 2] = xCurrent + xOffset;
circlePoints[10 * radius - doubleLoopIndex - 1] = -yCurrent + yOffset; // 这个是负的（对称）
楼主可以把 yOffset 和 xOffset 先设置为0，也即圆心就是原点，然后简化这个过程来想一想。