为什么Java代码加入小缓冲图像不工作?

Question

我创建了4个buffereimage对象，并用不同的颜色填充它们。然后创建一个更大的buffereimage对象，并绘制所有之前创建的四个buff。将图像放入其中以创建更大的图像。然后我将大图像保存到一个扩展名为gif的文件中。虽然大小是正确的，gif文件不包含颜色。为什么?我错在哪里?

import java.awt.HeadlessException;
import java.awt.*;
import java.awt.image.*;
import java.io.File;
import java.io.IOException;
import javax.imageio.ImageIO;
import javax.swing.JFrame;
import javax.swing.JPanel;
import javax.swing.WindowConstants;
public class MainFrame extends JFrame implements Runnable{
public MainFrame() throws HeadlessException {
ImagePanel ip=new ImagePanel(this);
add(ip,"Center");
}

public static void main(String[] args) {
MainFrame mf=new MainFrame();
mf.setDefaultCloseOperation(WindowConstants.DISPOSE_ON_CLOSE);
mf.setSize(new Dimension(800,600));
mf.setLocationRelativeTo(null);
mf.setVisible(true);
}
}
class ImagePanel extends JPanel {
public ImagePanel(MainFrame jf) {
ImageJoiner imj=new ImageJoiner();
}
}


class ImageJoiner {

// this will joing four buffered images together to construct bigger image as buffered image and save it as gif file

public ImageJoiner() {
createImages();
}

public void createImages() {

imtl=new BufferedImage(PW, PH, BufferedImage.TYPE_INT_RGB);
imtr=new BufferedImage(PW, PH, BufferedImage.TYPE_INT_RGB);
imbl=new BufferedImage(PW, PH, BufferedImage.TYPE_INT_RGB);
imbr=new BufferedImage(PW, PH, BufferedImage.TYPE_INT_RGB);

bigW=PW*2;
bigH=PH*2;

bigmap=new BufferedImage(bigW, bigH, BufferedImage.TYPE_INT_RGB); 

imtl.createGraphics().setColor(Color.blue);
imtl.createGraphics().fillRect(0,0, PW, PH);
imtr.getGraphics().setColor(Color.red);
imtr.getGraphics().fillRect(0, 0, PW, PH);
imbl.getGraphics().setColor(Color.green);
imbl.getGraphics().fillRect(0, 0, PW, PH);

imbr.getGraphics().setColor(Color.yellow);
imbr.getGraphics().fillRect(0, 0, PW, PH);

bigmap.getGraphics().drawImage(imtl, 0, 0,PW,PH,null);
bigmap.getGraphics().drawImage(imtr, PW, 0,PW,PH,null);
bigmap.getGraphics().drawImage(imbl, 0, PH, PW,PH,null);
bigmap.getGraphics().drawImage(imbr, PW, PH,PW,PH,null);

writeBufferedImageToFile(bigmap);

}

private void writeBufferedImageToFile(BufferedImage bim) {
File file = new File("images/out.gif");
try {
ImageIO.write(bim, "gif", file);
} catch (IOException e) {
// TODO Auto-generated catch block
e.printStackTrace();
}
}

static final int PW =1000;
static final int PH =1000;
static int bigW,bigH;
BufferedImage imtl,imtr,imbl,imbr;
BufferedImage bigmap;
}

Answer 1

您随意创建图形对象，设置一个对象的颜色，使用另一个对象进行绘图，这可能行不通。创建并以更好的方式使用图形对象，例如:

public static void createImages() {

imtl = new BufferedImage(PW, PH, BufferedImage.TYPE_INT_RGB);
imtr = new BufferedImage(PW, PH, BufferedImage.TYPE_INT_RGB);
imbl = new BufferedImage(PW, PH, BufferedImage.TYPE_INT_RGB);
imbr = new BufferedImage(PW, PH, BufferedImage.TYPE_INT_RGB);

int bigW = PW * 2;
int bigH = PH * 2;

BufferedImage bigmap = new BufferedImage(bigW, bigH, BufferedImage.TYPE_INT_RGB); 

Graphics2D g2d = imtl.createGraphics();
g2d.setColor(Color.BLUE);
g2d.fillRect(0,0, PW, PH);
g2d.dispose();
g2d = imtr.createGraphics();
g2d.setColor(Color.RED);
g2d.fillRect(0,0, PW, PH);
g2d.dispose();

g2d = imbl.createGraphics();
g2d.setColor(Color.GREEN);
g2d.fillRect(0,0, PW, PH);
g2d.dispose();
g2d = imbr.createGraphics();
g2d.setColor(Color.YELLOW);
g2d.fillRect(0,0, PW, PH);
g2d.dispose();

g2d = bigmap.createGraphics();
g2d.drawImage(imtl, 0, 0,PW,PH,null);
g2d.drawImage(imtr, PW, 0,PW,PH,null);
g2d.drawImage(imbl, 0, PH, PW,PH,null);
g2d.drawImage(imbr, PW, PH,PW,PH,null);
g2d.dispose();

writeBufferedImageToFile(bigmap);
}

经过测试的代码可以运行

---

您还问过:

我仍然很好奇，为什么BufferedImage#getGraphics().fillRect()不工作?它只工作你定义一个图形对象，然后分配它?为什么我不能直接使用它?

当BufferedImage#getGraphics()或BufferedImage#createGraphics()创建一个新的Graphics2D实例时，该实例与之前创建的任何实例都没有关系。

例如，您可以通过调用 进行测试:

BufferedImage img = new BufferedImage(20, 20, BufferedImage.TYPE_INT_RGB);
Graphics ga = img.getGraphics();
Graphics gb = img.getGraphics();
System.out.println("ga: " + ga.hashCode());
System.out.println("gb: " + gb.hashCode());
ga.dispose();  // always dispose of resources that *you* create
gb.dispose();

运行此程序，您将看到打印出两个不同的hashCodes，证明ga和gb实例是不同的。如果调用.createGraphics()，将出现相同的相对输出。

那么，当你这样做的时候:

imtl.createGraphics().setColor(Color.blue);
imtl.createGraphics().fillRect(0,0, PW, PH);

您正在设置一个图形实例的颜色，然后使用另一个独特的新实例进行绘图。由于没有设置第二个实例的颜色，它将使用默认颜色绘制，并且所有图像将保持黑色。