It is equal to counting leading zeros.
previous codes run in O(32).
this code (
http://codingways.blogspot.com/2012/06/count-leading-zero-bits-in-integer.html
) runs
in O(4)

Regards,
Hassan

On Thu, Jun 14, 2012 at 2:34 PM, utsav sharma <[email protected]>wrote:

> there are two approaches we can take
> one is of anika i.e. shifting a no. right till it becomes 0 and count no.
> of shifts
> int main()
> {
> int x=10;
> int cnt=-1;
> while(x)
> {
> x>>=1;
> cnt++;
> }
> printf("%d",cnt);
> }
>
> another is of aditya in which we didn't change the original no.
> instead take a new no. n=pow(2,15) for 16-bit and right shift it untill
> the first 1 occurs and displaying shift
> int main()
> {
> int x=10;
> int n=pow(2,15);
> int cnt=0;
> while(!(n & x))
> {
> n>>=1;
> cnt++;
> }
>
> printf("%d",16-(cnt+1));
> }
>
> On Thu, Jun 14, 2012 at 1:57 PM, aditya kumar <
> [email protected]> wrote:
>
>> for 16bit ..take a number n = 1000000000000000
>> let x be the number mentionf in a question
>> pos  = -1
>> while (1){
>> if(x & n){
>> pos++
>> exit}
>> else
>> n ยป=1
>> }
>>
>> pos is the answer
>> if pos is greater than numbr of bits then x doesnt hv set bits
>>
>>> How to find the LEFT MOST set bit in an unsigned integer.
>>> Example:
>>> "00*1*0 0001 0100 0010" is a 16-bit unsigned integer. We have to find
>>> the left most set bit position (Starting with 0 from right side )
>>> which is 13th position in our example. How to find this. Can any one pls
>>> give any suggestions.
>>> Thank You Very Much In Advance.
>>>
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>
>
>
> --
> Utsav Sharma,
> NIT Allahabad
>
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