What is the fastest way to count all set bits?

Question

What is the fastest way to count all set bits?
The number from which I am counting bits is from 1<=n<2^63, only positive values.

For now I use bitset from the library and it is pretty fast, but if there is any faster option I would like to know.

In the worst case scenario I have over 1 billion iterations, so I am looking a way to speed it up.
This is the part of my loop where I am counting the setted bits:

if (std::bitset<64>(currentNumber).count() == numofOnes)
++counter;

Answer 1

Here is a demo of what I mean with a balanced lookup table.
This one is heavy on the size saving.
You can adapt the concept to lookup tables of size e.g. 64 or 256, speeding up considerably.

#include <iostream>

using namespace std;

int main()
{
   unsigned long long int input=679043ULL; // just a big number, for demo
   
   unsigned char lookup[16]={0,1,1,2,1,2,2,3,1,2,2,3,2,3,3,4};
   unsigned char result=0;

   while (input>0)
   {
       cout << (unsigned int) result << " " << (input&15) << " " << input << endl;
       result+=lookup[input&15];
       input>>=4;
   }
   cout << (unsigned int)result << endl;
   return 0;
}

Output:

0 3 679043
2 8 42440
3 12 2652
5 5 165
7 10 10
9

The demo outputs shows that the loop accumulates
2 bits in "3",
1 bit in "8",
2 bits in "12",
2 bits in "5",
two bits in "10";
for a total of 9.