This is the code to reverse the bits in an unsigned integer .
Could anyone please explain the logic of this approach ? Thank You !!
#define reverse(x) \
(x=x16|(0xx)16, \
x=(0xff00ff00x)8|(0x00ff00ffx)8, \
x=(0xf0f0f0f0x)4|(0x0f0f0f0fx)4, \
x=(0xx)2|(0xx)2, \
x = x16 | (0xx)16
this line exchanges ls 16bits with ms 16bits, i.e. 1 pair of 16bit
this logic of exchanging bits is the used for 2 pairs of 8bits each, then
for 4 pairs of 4bit, then for 8 pairs of 2 bit and finally 16 pairs of 1bit.
On Fri, Aug 5, 2011 at 6:04 PM, rShetty
Hi Rajeev,
I follow similar approach. The basic logic is swap bits of a pair, then swap
nibbles(2 bits) and then swap (4bits), 8bits and go on.
So for ex. 0110 1101 1100 0101
In first step, I swap bits of each pair. So this becomes,
Input - 0110 1101 1100 0101
output- 1001 1110 1100 1010
In
@mithun : Thanks
On Fri, Aug 5, 2011 at 6:37 PM, mithun bs mithun...@gmail.com wrote:
Hi Rajeev,
I follow similar approach. The basic logic is swap bits of a pair, then
swap nibbles(2 bits) and then swap (4bits), 8bits and go on.
So for ex. 0110 1101 1100 0101
In first step, I swap bits
