## Tuesday, May 23, 2017

Suppose we have 16-bit number where we want to reverse some bits between bits i'th and j'th.  To illustrate that, say we have a 16-bit value stored in v:

|<-------- L = sizeof(short)*CHAR_BIT --------->|
15                       7                1   0
+--+--+--+--+--+--+--+--+--+--+--+--+--+--+--+--+
|  |  |  | x| x| x| x| x| x| x|  |  |  |  |  |  |
+--+--+--+--+--+--+--+--+--+--+--+--+--+--+--+--+
^                  ^
|                  |
i=12               j=6
|<------>|
L - i - 1

So, we want to reverse bits between bits 6th to 12th.

Assume i is always greater than j (i > j):
Number of bits to reverse: n = (i - j) + 1

Mask to use: (2^n - 1) << j

Or, in the case above:

n = (12 - 6) + 1 = 7
mask = (2^7 - 1) << 6 = 127 << 6  = 0x1FC0 (0001111111000000 in binary)

Then we mask the bits stored in v and shift it back to 0 (to do reverse operation):

r = (v & mask)  >> j;

r =
15                       7                1   0
+--+--+--+--+--+--+--+--+--+--+--+--+--+--+--+--+
| 0| 0| 0| 0| 0| 0| 0| 0| 0| x| x| x| x| x| x| x|
+--+--+--+--+--+--+--+--+--+--+--+--+--+--+--+--+

We use bit reversal operation as posted previously on r:

r = bit_reversal(r);

After this call, the bits are reversed, but filled in from MSB.  We need to put them to the position originally started.

15                       7                1   0
+--+--+--+--+--+--+--+--+--+--+--+--+--+--+--+--+
| x| x| x| x| x| x| x|  |  |  |  |  |  |  |  |  |
+--+--+--+--+--+--+--+--+--+--+--+--+--+--+--+--+
|-------> shr 3

<========= sizeof(unsigned int)* CHAR_BIT ======>

m = sizeof(short) * CHAR_BIT - n - j
m = 2*8 - 7 -6 = 16 - 13 = 3

Final result is: