What is the best practice way to create a bitmask for a range of bits?

Question

I can think of three ways to do this off the top of my head. I\'ll outline them real quick.

char mask = (1<>bot
m         


        

Answer 1

You could try a lookup table approach:

static const char LUT[][] = { // index like this LUT[bot][top]
//top:    0     1     2     3     4     5     6     7     8
       0x00, 0x01, 0x03, 0x07, 0x0F, 0x1F, 0x3F, 0x7F, 0xFF, // bot: 0
       0x00, 0x00, 0x02, 0x06, 0x0E, 0x1E, 0x3E, 0x7E, 0xFE, // bot: 1
       0x00, 0x00, 0x00, 0x04, 0x0C, 0x1C, 0x3C, 0x7C, 0xFC, // bot: 2
       0x00, 0x00, 0x00, 0x00, 0x00, 0x18, 0x38, 0x78, 0xF8, // bot: 3
       0x00, 0x00, 0x00, 0x00, 0x00, 0x10, 0x30, 0x70, 0xF0, // bot: 4
       0x00, 0x00, 0x00, 0x00, 0x00, 0x00, 0x20, 0x60, 0xE0, // bot: 5
       0x00, 0x00, 0x00, 0x00, 0x00, 0x00, 0x00, 0x40, 0xC0, // bot: 6
       0x00, 0x00, 0x00, 0x00, 0x00, 0x00, 0x00, 0x00, 0x80, // bot: 7
       0x00, 0x00, 0x00, 0x00, 0x00, 0x00, 0x00, 0x00, 0x00  // bot: 8
};

char mask = LUT[bot][top];

Also: If for whatever reason you go with bit manipulation this solution requires less ops. In addition a superscalar processor should evaluate the left and right side of the xor in parallel.

char mask = (0xFF << top) ^ (0xFF << bot);