@Dave Thank You very much :)
On Jun 27, 8:48 pm, piyush kapoor wrote:
> thanks a lot for the wonderful explanation :-)
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> On Mon, Jun 27, 2011 at 7:17 PM, Dave wrote:
> > @Rajeev and Piyush: Numbering the bits from the right starting with 0
> > as usual, you see that you need to
thanks a lot for the wonderful explanation :-)
On Mon, Jun 27, 2011 at 7:17 PM, Dave wrote:
> @Rajeev and Piyush: Numbering the bits from the right starting with 0
> as usual, you see that you need to move the even-numbered bits one bit
> to the left and the odd-numbered bits one bit to the righ
@Rajeev and Piyush: Numbering the bits from the right starting with 0
as usual, you see that you need to move the even-numbered bits one bit
to the left and the odd-numbered bits one bit to the right. You could
do this one bit-pair at a time, but it would be more efficient if you
could do all pairs
Yep,I also want to know the same..
On Mon, Jun 27, 2011 at 6:23 PM, rajeev bharshetty wrote:
> @ Dave How to think about the answer to the above question . I mean How do
> I tackle such problems ?
>
>
> On Mon, Jun 27, 2011 at 6:17 PM, Dave wrote:
>
>> y = ((x & 0x55) << 1) | ((x >> 1) & 0x55).
@ Dave How to think about the answer to the above question . I mean How do I
tackle such problems ?
On Mon, Jun 27, 2011 at 6:17 PM, Dave wrote:
> y = ((x & 0x55) << 1) | ((x >> 1) & 0x55).
>
> Note, 0x55 = 01010101 in binary.
>
> Dave
>
> On Jun 27, 7:18 am, rShetty wrote:
> > Given a byte, wr
y = ((x & 0x55) << 1) | ((x >> 1) & 0x55).
Note, 0x55 = 01010101 in binary.
Dave
On Jun 27, 7:18 am, rShetty wrote:
> Given a byte, write a code to swap every two bits. [Using bit
> operators]
>
> Eg: Input: 10 01 11 01 Output: 01 10 11 10
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