Give a very good method to count the number of ones in a "n" (e.g. 3

Give a very good method to count the number of ones in a "n"
(e.g. 32) bit number.

Question Posted / sujan

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Give a very good method to count the number of ones in a "n" (e.g. 32) bit number...

Answer / sujan

#include<iostream>
#define bit 32
using namespace std;

int array[bit];

int bitConvert(int n)
{
int a,j=0;

a=n%2;
for(int i=bit;i>=0;i--)
{
n=n/2;
array[i]=a;
a=n%2;
}
for(int i=0;i<=bit;i++)
{
cout<<array[i];

}
}
int countBit(int a[])
{
int *ptr;
ptr=a;
int j=0;
for(int i=0;i<=bit;i++)
{

if(*ptr==1)
{
j++;
}
ptr++;
}
cout<< j;
}

int main()
{
int n;
cout<<"Enter the no:";
cin>>n;
cout<<"\n"<<"BitConversion of "<<n<< "is:";
bitConvert(n);
cout<<endl<<endl;
cout<<"\n"<<"No. of bit:";
countBit(array);
system("pause");
}

Is This Answer Correct ?

4 Yes

1 No

Give a very good method to count the number of ones in a "n" (e.g. 32) bit number...

Answer / artyom

// It's still O(n), maybe there are better ways.
int countBit(int num)
{
int count = 0;
while(num)
{
count += static_cast<int>(static_cast<bool>(mask &
0x1));
num >>= 1;
}
return count;
}

Is This Answer Correct ?

2 Yes

0 No

Give a very good method to count the number of ones in a "n" (e.g. 32) bit number...

Answer / manish kumar

for (c = 0; n; c++)
n &= n - 1;

Result: the value of c.

complexity:o(logn)

Is This Answer Correct ?

1 Yes

0 No

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