Question

我想将RGB颜色编码为单个整数值。

假设编码算法是这样的：

int code = (blue * 256 * 256) + (green * 256) + red

如何使用位移和/或按位运算符对代码中的RGB组件进行编码/解码？

Answer 1

    int blueMask = 0xFF0000, greenMask = 0xFF00, redMask = 0xFF;
    int r = 12, g = 13, b = 14;
    int bgrValue = (b << 16) + (g << 8) + r;
    System.out.println("blue:" + ((bgrValue & blueMask) >> 16));
    System.out.println("red:" + ((bgrValue & redMask)));
    System.out.println("green:" + ((bgrValue & greenMask) >> 8));

Answer 2

如果您只想进行RGB转换，而不关心我建议使用java.awt.Color

int r = 255; //red
int g = 255; //green
int b = 255; //blue
int a = 255; //alpha
Color c = new Color(r,g,b,a);

使用getRGB方法和getRed，getBlue，getGreen方法

int RGB = c.getRGB();
int red = c.getRed();
int blue = c.getBlue();
int green = c.getGreen();

或者，您可以使用Color(r,g,b)构造函数构造颜色对象，它将具有默认的255 alpha。

使用位操作（ARGB，32位色彩空间）。构造RGB颜色：

int alpha = 255;    
int red = 128;
int green = 128;
int blue = 128;
int RGB = (alpha << 24);
RGB = RGB | (red << 16);
RGB = RGB | (green << 8);
RGB = RGB | (blue);

System.out.println(Integer.toBinaryString(RGB));

出11111111100000001000000010000000

解码按照评论中的链接完成。

Answer 3

这是一个我做过的模拟程序，可以帮助你。我接下来的转换很像Dev Blanked基于我做过的旧程序，但他在我把程序放在一起时回答。因为我还是做了这项工作，想想我会分享它以任何方式帮助它。

import java.util.Scanner;
import java.math.*;

public class RGB{

public static void main(String[]args){
    Scanner scan = new Scanner(System.in);
    int code; //Code for the color
    int red, green, blue; //Individual colors
    int rMask = 0xFF0000, gMask = 0xFF00, bMask = 0xFF; //Masks for the colors

    //Take input
    System.out.println("Please enter the red color. Range [0, 255] only please.");
    red = scan.nextInt();
    System.out.println("Please enter the green color. Range [0, 255] only please.");
    green = scan.nextInt();
    System.out.println("Please enter the blue color. Range [0, 255] only please.");
    blue = scan.nextInt();

    //Generate code based on Behnil's way.
    code = 0;
    code += (int) (red * Math.pow(2, 16));
    code += (int) (green * Math.pow(2, 8));
    code += (int) (blue * Math.pow(2,0));
    System.out.println("The code is " + code + ".");

    //Clear values
    red = 0;
    green = 0;
    blue = 0;

    //Obtain values.
    red = (code & rMask) >> 16;
    green = (code & gMask) >> 8;
    blue = (code & bMask);

    System.out.println("Your red value is: " + red);
    System.out.println("Your green value is: " + green);
    System.out.println("Your blue value is: " + blue);      
}

}

Answer 4

public static void main(String[] args){
    int red = 111;
    int green = 222;
    int blue = 121;

    int code = red*256*256 + green*256 + blue;

    blue = code%256;
    green = (code%(256*256) - blue)/256;
    red = (code - blue - green*256)/(256*256); 

    System.out.println("" + red + green + blue);
}

按预期输出111222121。这是我修复它的方式，但我不确定专业人士是否同意这一点，因为它可能比使用位移更慢

位移和按位运算可对RGB值进行编码

4 个答案: