How to count the number of set bits in a 32-bit in

Question

8 bits representing the number 7 look like this:

00000111

Three bits are set.

What are algorithms to determine the number of set bits in a 32-bit integer?

Answer 1

I use the below code which is more intuitive.

int countSetBits(int n) {
    return !n ? 0 : 1 + countSetBits(n & (n-1));
}

Logic : n & (n-1) resets the last set bit of n.

P.S : I know this is not O(1) solution, albeit an interesting solution.

Answer 2

In my opinion, the "best" solution is the one that can be read by another programmer (or the original programmer two years later) without copious comments. You may well want the fastest or cleverest solution which some have already provided but I prefer readability over cleverness any time.

unsigned int bitCount (unsigned int value) {
    unsigned int count = 0;
    while (value > 0) {           // until all bits are zero
        if ((value & 1) == 1)     // check lower bit
            count++;
        value >>= 1;              // shift bits, removing lower bit
    }
    return count;
}

If you want more speed (and assuming you document it well to help out your successors), you could use a table lookup:

// Lookup table for fast calculation of bits set in 8-bit unsigned char.

static unsigned char oneBitsInUChar[] = {
//  0  1  2  3  4  5  6  7  8  9  A  B  C  D  E  F (<- n)
//  =====================================================
    0, 1, 1, 2, 1, 2, 2, 3, 1, 2, 2, 3, 2, 3, 3, 4, // 0n
    1, 2, 2, 3, 2, 3, 3, 4, 2, 3, 3, 4, 3, 4, 4, 5, // 1n
    : : :
    4, 5, 5, 6, 5, 6, 6, 7, 5, 6, 6, 7, 6, 7, 7, 8, // Fn
};

// Function for fast calculation of bits set in 16-bit unsigned short.

unsigned char oneBitsInUShort (unsigned short x) {
    return oneBitsInUChar [x >>    8]
         + oneBitsInUChar [x &  0xff];
}

// Function for fast calculation of bits set in 32-bit unsigned int.

unsigned char oneBitsInUInt (unsigned int x) {
    return oneBitsInUShort (x >>     16)
         + oneBitsInUShort (x &  0xffff);
}

Although these rely on specific data type sizes so they're not that portable. But, since many performance optimisations aren't portable anyway, that may not be an issue. If you want portability, I'd stick to the readable solution.

Answer 3

Few open questions:-

1. If the number is negative then?
2. If the number is 1024 , then the "iteratively divide by 2" method will iterate 10 times.

we can modify the algo to support the negative number as follows:-

count = 0
while n != 0
if ((n % 2) == 1 || (n % 2) == -1
    count += 1
  n /= 2  
return count

now to overcome the second problem we can write the algo like:-

int bit_count(int num)
{
    int count=0;
    while(num)
    {
        num=(num)&(num-1);
        count++;
    }
    return count;
}

for complete reference see :

http://goursaha.freeoda.com/Miscellaneous/IntegerBitCount.html

Answer 4

I think the fastest way—without using lookup tables and popcount—is the following. It counts the set bits with just 12 operations.

int popcount(int v) {
    v = v - ((v >> 1) & 0x55555555);                // put count of each 2 bits into those 2 bits
    v = (v & 0x33333333) + ((v >> 2) & 0x33333333); // put count of each 4 bits into those 4 bits  
    return c = ((v + (v >> 4) & 0xF0F0F0F) * 0x1010101) >> 24;
}

It works because you can count the total number of set bits by dividing in two halves, counting the number of set bits in both halves and then adding them up. Also know as Divide and Conquer paradigm. Let's get into detail..

v = v - ((v >> 1) & 0x55555555); 

The number of bits in two bits can be 0b00, 0b01 or 0b10. Lets try to work this out on 2 bits..

 ---------------------------------------------
 |   v    |   (v >> 1) & 0b0101   |  v - x   |
 ---------------------------------------------
   0b00           0b00               0b00   
   0b01           0b00               0b01     
   0b10           0b01               0b01
   0b11           0b01               0b10

This is what was required: the last column shows the count of set bits in every two bit pair. If the two bit number is >= 2 (0b10) then and produces 0b01, else it produces 0b00.

v = (v & 0x33333333) + ((v >> 2) & 0x33333333); 

This statement should be easy to understand. After the first operation we have the count of set bits in every two bits, now we sum up that count in every 4 bits.

v & 0b00110011         //masks out even two bits
(v >> 2) & 0b00110011  // masks out odd two bits

We then sum up the above result, giving us the total count of set bits in 4 bits. The last statement is the most tricky.

c = ((v + (v >> 4) & 0xF0F0F0F) * 0x1010101) >> 24;

Let's break it down further...

v + (v >> 4)

It's similar to the second statement; we are counting the set bits in groups of 4 instead. We know—because of our previous operations—that every nibble has the count of set bits in it. Let's look an example. Suppose we have the byte 0b01000010. It means the first nibble has its 4bits set and the second one has its 2bits set. Now we add those nibbles together.

0b01000010 + 0b01000000

It gives us the count of set bits in a byte, in the first nibble 0b01100010 and therefore we mask the last four bytes of all the bytes in the number (discarding them).

0b01100010 & 0xF0 = 0b01100000

Now every byte has the count of set bits in it. We need to add them up all together. The trick is to multiply the result by 0b10101010 which has an interesting property. If our number has four bytes, A B C D, it will result in a new number with these bytes A+B+C+D B+C+D C+D D. A 4 byte number can have maximum of 32 bits set, which can be represented as 0b00100000.

All we need now is the first byte which has the sum of all set bits in all the bytes, and we get it by >> 24. This algorithm was designed for 32 bit words but can be easily modified for 64 bit words.

Answer 5

Also consider the built-in functions of your compilers.

On the GNU compiler for example you can just use:

int __builtin_popcount (unsigned int x);
int __builtin_popcountll (unsigned long long x);

In the worst case the compiler will generate a call to a function. In the best case the compiler will emit a cpu instruction to do the same job faster.

The GCC intrinsics even work across multiple platforms. Popcount will become mainstream in the x86 architecture, so it makes sense to start using the intrinsic now. Other architectures have the popcount for years.

---

On x86, you can tell the compiler that it can assume support for popcnt instruction with -mpopcnt or -msse4.2 to also enable the vector instructions that were added in the same generation. See GCC x86 options. -march=nehalem (or -march= whatever CPU you want your code to assume and to tune for) could be a good choice. Running the resulting binary on an older CPU will result in an illegal-instruction fault.

To make binaries optimized for the machine you build them on, use -march=native (with gcc, clang, or ICC).

MSVC provides an intrinsic for the x86 popcnt instruction, but unlike gcc it's really an intrinsic for the hardware instruction and requires hardware support.

---

Using std::bitset<>::count() instead of a built-in

In theory, any compiler that knows how to popcount efficiently for the target CPU should expose that functionality through ISO C++ std::bitset<>. In practice, you might be better off with the bit-hack AND/shift/ADD in some cases for some target CPUs.

For target architectures where hardware popcount is an optional extension (like x86), not all compilers have a std::bitset that takes advantage of it when available. For example, MSVC has no way to enable popcnt support at compile time, and always uses a table lookup, even with /Ox /arch:AVX (which implies SSE4.2, although technically there is a separate feature bit for popcnt.)

But at least you get something portable that works everywhere, and with gcc/clang with the right target options, you get hardware popcount for architectures that support it.

#include <bitset>
#include <limits>
#include <type_traits>

template<typename T>
//static inline  // static if you want to compile with -mpopcnt in one compilation unit but not others
typename std::enable_if<std::is_integral<T>::value,  unsigned >::type 
popcount(T x)
{
    static_assert(std::numeric_limits<T>::radix == 2, "non-binary type");

    // sizeof(x)*CHAR_BIT
    constexpr int bitwidth = std::numeric_limits<T>::digits + std::numeric_limits<T>::is_signed;
    // std::bitset constructor was only unsigned long before C++11.  Beware if porting to C++03
    static_assert(bitwidth <= std::numeric_limits<unsigned long long>::digits, "arg too wide for std::bitset() constructor");

    typedef typename std::make_unsigned<T>::type UT;        // probably not needed, bitset width chops after sign-extension

    std::bitset<bitwidth> bs( static_cast<UT>(x) );
    return bs.count();
}

See asm from gcc, clang, icc, and MSVC on the Godbolt compiler explorer.

x86-64 gcc -O3 -std=gnu++11 -mpopcnt emits this:

unsigned test_short(short a) { return popcount(a); }
    movzx   eax, di      # note zero-extension, not sign-extension
    popcnt  rax, rax
    ret
unsigned test_int(int a) { return popcount(a); }
    mov     eax, edi
    popcnt  rax, rax
    ret
unsigned test_u64(unsigned long long a) { return popcount(a); }
    xor     eax, eax     # gcc avoids false dependencies for Intel CPUs
    popcnt  rax, rdi
    ret

PowerPC64 gcc -O3 -std=gnu++11 emits (for the int arg version):

    rldicl 3,3,0,32     # zero-extend from 32 to 64-bit
    popcntd 3,3         # popcount
    blr

This source isn't x86-specific or GNU-specific at all, but only compiles well for x86 with gcc/clang/icc.

Also note that gcc's fallback for architectures without single-instruction popcount is a byte-at-a-time table lookup. This isn't wonderful for ARM, for example.

Answer 6

For a happy medium between a 2³² lookup table and iterating through each bit individually:

int bitcount(unsigned int num){
    int count = 0;
    static int nibblebits[] =
        {0, 1, 1, 2, 1, 2, 2, 3, 1, 2, 2, 3, 2, 3, 3, 4};
    for(; num != 0; num >>= 4)
        count += nibblebits[num & 0x0f];
    return count;
}

From http://ctips.pbwiki.com/CountBits