Java 检索到的值具有空值。如何消除它们？

我已经成功地从CompareExargb方法中检索到RGB像素值。而且输出似乎不令人信服，因为输出也包含一个空字符。我怎样才能消除它们

其次，当我尝试组合字符串[][]char 1+char 2时，我得到一个错误，指出它是“+”类型的错误操作数。但是我想合并2个字符串，以便能够将其转换为ASCII字符

下面是compareHexaRGB方法的代码：

public class compareHexaRGB
{
 private static int w;
 private static int h;
 private static BufferedImage img;
 private static BufferedImage img2;
 private static String[][] check_hex2;
 private static String[][] check_hex4;
 private static String[][] message;

 public static void compareHexaRGB(BufferedImage image, BufferedImage image2, int width, int height) throws IOException
{
w = width;
h = height;
img = image;
img2 = image2;

}

public void check() throws IOException
{
    getPixelRGB1 pixel = new getPixelRGB1();
    getPixelData1 newPD = new getPixelData1();

    int[] rgb;
    int count = 0;

    int[][] pixelData = new int[w * h][3];
    check_hex2 = new String[w][h];
    check_hex4 = new String[w][h];

    for(int i = 0; i < w; i++) 
    {
        for(int j = 0; j < h; j++)
        {
            rgb = newPD.getPixelData(img, i, j);

            for(int k = 0; k < rgb.length; k++)
            {
                pixelData[count][k] = rgb[k];
            }

                if(pixel.display_imgHex2()[i][j].equals(pixel.display_img2Hex2()[i][j]))
                {
                    System.out.println("\nPixel values at position 2 are the same." + "\n" + pixel.display_imgHex2()[i][j] + "  " + pixel.display_img2Hex2()[i][j]);
                }
                if(pixel.display_imgHex4()[i][j].equals(pixel.display_img2Hex4()[i][j]))
                {
                    System.out.println("\nPixel values at position 4 are the same." + "\n" + pixel.display_imgHex4()[i][j] + "  " + pixel.display_img2Hex4()[i][j]);
                }
                if(!pixel.display_imgHex2()[i][j].equals(pixel.display_img2Hex2()[i][j]))
                {
                    System.out.println("\nPixel values at position 2 are not the same." + "\n" + pixel.display_imgHex2()[i][j] + "  " + pixel.display_img2Hex2()[i][j]);
                    check_hex2[i][j] = pixel.display_img2Hex2()[i][j];
                    System.out.println("\nOutput Hex 2: " + check_hex2[i][j]);
                }
                if(!pixel.display_imgHex4()[i][j].equals(pixel.display_img2Hex4()[i][j]))
                {
                    System.out.println("\nPixel values at position 4 are not the same." + "\n" + pixel.display_imgHex4()[i][j] + "  " + pixel.display_img2Hex4()[i][j]);
                    check_hex4[i][j] = pixel.display_img2Hex4()[i][j];
                    System.out.println("\nOutput Hex 4: " + check_hex4[i][j]);
                }
                if(!pixel.display_imgHex2()[i][j].equals(pixel.display_img2Hex2()[i][j]) || (!pixel.display_imgHex4()[i][j].equals(pixel.display_img2Hex4()[i][j])))
                {
                    System.out.println("\nOne of the pixel values at position 2 and 4 are not the same." + "\n" + pixel.display_imgHex2()[i][j] + "  " + pixel.display_img2Hex2()[i][j] + "\n" + pixel.display_imgHex4()[i][j] + "  " + pixel.display_img2Hex4()[i][j]);

                    if(!pixel.display_imgHex2()[i][j].equals(pixel.display_img2Hex2()[i][j]) || (pixel.display_imgHex2()[i][j].equals(pixel.display_img2Hex2()[i][j])))
                    {
                        check_hex2[i][j] = pixel.display_img2Hex2()[i][j];
                        System.out.println("\nOutput Hex 2: " + check_hex2[i][j]);
                    }

                    if(!pixel.display_imgHex4()[i][j].equals(pixel.display_img2Hex4()[i][j]) || (pixel.display_imgHex4()[i][j].equals(pixel.display_img2Hex4()[i][j])))
                    {
                        check_hex4[i][j] = pixel.display_img2Hex4()[i][j];
                        System.out.println("\nOutput Hex 4: " + check_hex4[i][j]);
                    }

                }
            count++;
            System.out.println("\nOutput Count: " + count);
        }

    }

}

public String[][] getCheck_hex2()
{
    return check_hex2;
}

public String[][] getCheck_hex4()
{
    return check_hex4;
}
}

如有任何建议或更正，我将不胜感激

• 由于这些位置的像素是相同的，所以会得到很多空值。在

  check（）

  方法中，当像素相同时，不向该位置的数组中添加任何内容，因此它将保持为空。在这些点上添加一个放置字符，如

  -

  或其他

• 您的打印方法是在一行上打印每个元素，如果您像打印矩阵一样打印二维数组，则可读性会更好。您可以通过将方法修改为

  for (int i = 0; i < inn.length; i++) 
  {
      for(int j = 0; j < inn[i].length; j++)
      {
         System.out.print(inn[i][j] + " ");
      }
      System.out.println();
  }  
  

• 当您尝试添加数组时会出现错误，因为您根本无法做到这一点-在不循环内容的情况下将两个数组添加到一起。您将需要一个与上面类似的循环

  char1 = hexRGB.getCheck_hex2();
  char2 = hexRGB.getCheck_hex4();
  //assuming char1 and char2 are the same size and symmetrical
  String[][] combine = new String[char1.length][char1[0].length]
  
  for (int i = 0; i < char1.length; i++) 
  {
      for(int j = 0; j < char1[i].length; j++)
      {
           // This will only concatenate the strings though,
           // If you want the numerical addition you will have to 
           // convert them to ints first
           combine[i][j] = char1[i][j] + char2[i][j];
      }
  }
  

  char1=hexRGB.getCheck_hex2（）；
  char2=hexRGB.getCheck_hex4（）；
  //假设char1和char2大小相同且对称
  字符串[][]组合=新字符串[char1.length][char1[0].length]
  for（int i=0；i

---

在您的第一个建议中，您表示我应该添加一个放置角色的位置。我可以这样做吗-->如果（pixel.display\u imgHex2（）[i][j].equals（pixel.display\u img2Hex2（）[i][j]）{System.out.println（“\n位置2处的像素值相同。”+“\n”+pixel.display\u imgHex2（）[i][j]+”+pixel.display\u img2Hex2（）[i][j]）；而不是stega2[i][j]=pixel.display_img2Hex2（）[i][j]；}？还是我做错了？

for (int i = 0; i < inn.length; i++) 
{
    for(int j = 0; j < inn[i].length; j++)
    {
       System.out.print(inn[i][j] + " ");
    }
    System.out.println();
}  

for(String[] arr : inn)
     System.out.println(Arrays.toString(arr));

char1 = hexRGB.getCheck_hex2();
char2 = hexRGB.getCheck_hex4();
//assuming char1 and char2 are the same size and symmetrical
String[][] combine = new String[char1.length][char1[0].length]

for (int i = 0; i < char1.length; i++) 
{
    for(int j = 0; j < char1[i].length; j++)
    {
         // This will only concatenate the strings though,
         // If you want the numerical addition you will have to 
         // convert them to ints first
         combine[i][j] = char1[i][j] + char2[i][j];
    }
}