自行计算的 RGB 值与实际 RGB 值不匹配

Question

我正在开发一个程序，我在其中获取图像的一部分并计算该图像的平均 RGB。当我计算它时，我得到完全不同的值，就好像我要使用内置函数一样。当我测试我的数字并将它们放入 RGB 颜色图表时，它们是关闭的，而内置功能非常准确。使用内置函数的问题在于它实际上没有任何功能。它只是打印出我计算的内容，但我无法使用该数据。 link to function called dump()

public void readSquares(Mat img){

    int width = 20;
    int height = 20;
    int rSum = 0;
    int gSum = 0;
    int bSum = 0;
    int rAvg = 0;
    int gAvg = 0;
    int bAvg = 0;

    Imgproc.cvtColor(img, img, Imgproc.COLOR_BGR2RGB);   
    int channels = img.channels();

    int totalBytes = (int)(img.total() * img.channels());

    byte buff[] = new byte[totalBytes];
    img.get(0, 0, buff);

    for (int i=0; i< height; i++) {
        //stride is the number of bytes in a row of smallImg
        int stride = channels * width;
        for (int j=0; j<stride; j+=channels) {
            int r = buff[(i * stride) + j];
            int g = buff[(i * stride) + j + 1]; 
            int b = buff[(i * stride) + j + 2];

            if(r < 0 || g < 0 || b < 0){
                r = Math.abs(r);
                g = Math.abs(g);
                b = Math.abs(b);    
            }

            rSum += r;
            gSum += g;
            bSum += b;

        }
    }   

    rAvg = (int) (rSum / img.total()); //total pixels in picture
    gAvg = (int) (gSum / img.total());
    bAvg = (int) (bSum / img.total());
    System.out.println("R: " + rAvg);
    System.out.println("G: " + gAvg);
    System.out.println("B: " + bAvg);

下面会打印出准确的值，但我无法获得单独的通道或找到平均 RGB，它只是将所有内容作为我无法使用的字符串吐出。

System.out.println(img.dump());
}

这是一个例子；我为像素读取了红色图像，这是输出：

我的计算

R:18
G:27
B:64

功能 第一个数字代表 R，第二个是 G，第三个是 B。它对图像中的每个像素都执行此操作。

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Answer 1

严格来说，Java 语言规范规定字节数据类型的范围为 -128 - 127，这就是 Java 解释字节的方式。例如，如果一个字节被提升或强制转换为 int Java 将解释第一位作为符号并使用符号扩展。没有语言支持可以帮助您以不同的方式解释字节。 （例如，Java 中没有无符号关键字。）

话虽如此，没有什么能阻止您将字节视为 8 位值并将这些位解释为无符号位。请记住，您无法将自己的解释强加于其他人的方法。如果一个方法接受一个字节，那么该方法接受一个介于 -128 和 127 之间的值，除非另有明确说明。

    public static int unsignedToBytes(byte b) {
        return b & 0xFF;
    }

此方法可以满足您的要求。希望这会有所帮助！！！

编辑：在 Stack Overflow 和其他地方有很多好的答案。我建议您在发布问题之前尝试找到答案。谢谢。