we can use masking of most significant bit of the xor of (a-x) and ( b-x)
to check if a<x<b
if [ ((a-x)^(b-x)) & 0x80000000 ]
then it's in range
else not in range
On Sat, Oct 27, 2012 at 11:17 PM, Vikram Pradhan <[email protected]>wrote:
> how about ...if( (a-x)^(b-x) < 0)
>
>
>
>
> On Sat, Oct 27, 2012 at 9:21 PM, Dave <[email protected]> wrote:
>
>> @Atul: Try x = 0, a = 1, b = 2, for which (x > a && x < b) is false, but
>> (x - a < b - x) is true.
>>
>> Dave
>>
>> On Saturday, October 27, 2012 5:43:17 AM UTC-5, ATul SIngh wrote:
>>
>>> Thnks but i figured it out
>>> as
>>> ( x >= a && x < b )
>>> or
>>> ( x >= a && x <= b )
>>>
>>> could be written as
>>> x -a < b - x for first case
>>> x -a <= b - x for sec one
>>>
>>> Second set is more optimised as it contains only one comparison and
>>> minus is much efficient than comparison operator.
>>>
>>>
>>>
>>> --
>>>
>>> *ATul Singh** Software Engineer**, Interra Systems*
>>> Mobile : +91-9971794013
>>> www.interrrasystems.com
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>>>
>>>
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>
>
>
> --
> Vikram Pradhan | B.Tech| Computer Science & Engineering | NIT Jalandhar |
> 9740186063 |
>
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