确实与Java像素操作混淆

Question

THIS part i'm confused with what these 3 hexadecimals are doing such as 00ff0000, and why they're bitshifted 16 or 8 places

 // Getting pixel color by position x=100 and y=40 
    int clr=  image.getRGB(100,40); 
    int  red   = (clr & 0x00ff0000) >> 16;
    int  green = (clr & 0x0000ff00) >> 8;
    int  blue  =  clr & 0x000000ff;

Answer 1

The pixel's color information is encoded in a single 32 bit integer. The lowest eight bits store the blue color information, bits 8 to 15 store green and 16 to 23 store red. Bits 24 to 31 store the alpha value. The code that you show first selects the right bits by masking them using the and operation. In order to do calculations with them they are moved to represent their actual values.

clr & 0x0000ff00

selects the bits at position 8 to 15,

(clr & 0x0000ff00) >> 8

moves the result by 8 positions to the right.

Answer 2

In

int  red   = (clr & 0x00ff0000) >> 16;

The & will zero out all bits except those that are wanted:

0x00123456 & 0x00ff0000 == 0x00120000

The bit shift will place these bits at the desired position:

0x00120000 >> 16 == 0x00000012 == 0x12

Similarly for the other two channels.

0x00123456 & 0x0000ff00 == 0x00003400
0x00003400 >> 16 == 0x34

0x00123456 & 0x000000ff == 0x56

The reason for this is that the ARGB format stuffs four bytes (alpha, red, green, blue) into one int: 0xAaRrGgBb . The RGB format is similar except it doesn't use the alpha (opacity) channel. The whole point behind the bit-shifting is to separate those bytes: clr == 0x123456 to red == 0x12 green == 0x34 blue == 0x56

Note that each byte (8 bits) is represented by two hexadecimal digits (4 bits each) in the hexadecimal notation, so shifting by 16 bits shifts by 4*4 bits = 4 hexadecimal digits = 2 bytes.