it is
3*2^12 + 15*2^8 + 3*2^4 + 3
so it becomes
3<<12 + 15<<8 + 3<<4 +3.. let this equal to n
take a int count=0;
while(n){
n=n&(n-1);
count++;
}
return count;
2010/8/4 Seçkin Can Şahin <seckincansa...@gmail.com>
> either count it with a for loop or if you are using C, use
> __builtin_popcount(int x)
>
> or
>
> x is given
> int count = 0;
> for(int i=0; i<32;i++) count += (x & (1<<i)) != 0;
>
>
> On Tue, Aug 3, 2010 at 12:04 PM, amit <amitjaspal...@gmail.com> wrote:
>
>> (3*4096+15*256+3*16+3). How many 1's are there in the binary
>> representation of the result.
>>
>> Is there a quick way to count the number of set bits in this number
>> manually???
>>
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With Regards,
Jalaj Jaiswal
+919026283397
B.TECH IT
IIIT ALLAHABAD
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