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
<aditya.kumar130...@gmail.com>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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