Java 用于对 RGB 值进行编码的位移位和按位运算
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Bit Shift and Bitwise operations to encode RGB values
提问by Behnil
I would like to encode the RGB color to a single integer value.
我想将 RGB 颜色编码为单个整数值。
Let's say the algorithm for encoding is like this:
假设编码算法是这样的：
int code = (blue * 256 * 256) + (green * 256) + red
How can encode/decode RGB components to/from the code using bit shift and/or bitwise operators?
如何使用位移和/或按位运算符将 RGB 分量编码/解码为代码？
采纳答案by Dev Blanked
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));
回答by arynaq
If you simply want to/from RGB conversion and don't care how I would suggest using java.awt.Color
如果您只是想进行 RGB 转换并且不关心我建议如何使用 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);
The using the getRGBmethod and getRed, getBlue, getGreen methods
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();
Alternatively you can construct a color object using the Color(r,g,b)constructor, it will have default 255 alpha.
或者，您可以使用构造函数构造一个颜色对象Color(r,g,b)，它将具有默认的 255 alpha。
With bit operations (ARGB, 32 bit colorspace). Constructing the RGB color:
使用位操作（ARGB，32 位色彩空间）。构建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));
Out 11111111100000001000000010000000
出去 11111111100000001000000010000000
Decoding is done as in the link in the comment.
解码是在评论中的链接中完成的。
回答by Fiki
Here is a mock program that I've done up that might assist you. I approached the conversion much like Dev Blanked based off an old program I did, but he answered while I was putting the program together. Since I did the work anyways, figured I'd share in case it helped in any way.
这是我完成的一个模拟程序，可能会对您有所帮助。我根据我做的一个旧程序来处理转换，就像 Dev Blanked 一样，但他在我把程序放在一起时回答了。由于无论如何我都做了这项工作，我想我会分享以防它以任何方式提供帮助。
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);
}
}
回答by iLikeCookiesQ
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);
}
this outputs 111222121 as intended. This is the way i fixed it but i am not sure if the pro's agree with this as it might be slower than using bitshifts
这将按预期输出 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，因为它可能比使用移位慢
