使用C ++按位运算符查找匹配的颜色

Question

I want to figure out whether jersey1 has a Blue pixel in it or not. Also, I have find this using bitwise operator. For doing so, should I use the bitwise and (&) or bitwise or (|). Can you please explain me how they differ from each other? Thanks in advance.

 const unsigned int GREEN= 0x0000FF00;  //green
 const unsigned int RED  = 0x00FF0000;    //red
 const unsigned int BLUE = 0x000000FF;   //blue
 const unsigned int RGB  = RED | GREEN | BLUE;

 unsigned int jersey1 = 123;

Answer 1

You use the bitwise and operator & to mask the values you are interested in.

For example, to see if an blue bits are set you would say jersey1 & BLUE . If the value is 0 then there are no blue bits. If it's non-zero then there is a blue bit set.

Answer 2

You'll want to use a bitwise '&' to test if a blue pixel exists in jersey1 like the following:

if(jersey1 & BLUE)
{
   cout << "It's blue!" << endl;
}
else
{
   cout << "It's not blue!" << endl;
}

The '&' or "and" operator with the bits requires that both bits in the same position of each value (BLUE and jersey1) are equal to 1 for it to be true. Whereas, the '|' (or) operator only requires that one set of bits in one of the values is equal to 1. In the above example, had you used "bitwise or" (|), the if statement would always be true which is not what you want.

Answer 3

You can think of the bitwise AND and OR operators as gates that either stop values (mask them out) or let them through. If you take a value and a mask, and compare their binary representations, then the AND operator will have a 1 anywhere that both have a 1. The OR operator will have a 1 wherever at least one of them has a 1. It looks like this, reading vertically:

0 1 0 1
0 0 1 1
-------  AND
0 0 0 1

0 1 0 1
0 0 1 1
------- OR
0 1 1 1

Colors are frequently represented by hexadecimal values which contain 256 possible shades of each of the red, green, and blue components of the color. The bytes are (usually) ordered RRGGBB. Therefore you should convert your int 123 to hexadecimal to get 0x0000007B and then choose a mask which will determine the value that's in the "blue" position. {Actually, we can already see that there is a value in the blue position, so we could stop now...} Anyway, to simply mask a value down to its blue channel, we would AND with a mask which only passes through the values in that last position. Since FF in hex gives 11111111 and ones pass values through, we can mask with 0000FF , which is the mask you were given for the blue channel (the BLUE const). You'd simply need to test the value of (jersey1 & BLUE) to see if it's nonzero to find out whether there's blue in that channel.