Setting a bit
Use the bitwise OR operator (
|) to set a bit.
number |= 1UL << n;
That will set the
nth bit of
n should be zero, if you want to set the
1st bit and so on upto
n-1, if you want to set the
number is wider than
unsigned long; promotion of
1UL << n doesn't happen until after evaluating
1UL << n where it's undefined behaviour to shift by more than the width of a
long. The same applies to all the rest of the examples.
Clearing a bit
Use the bitwise AND operator (
&) to clear a bit.
number &= ~(1UL << n);
That will clear the
nth bit of
number. You must invert the bit string with the bitwise NOT operator (
~), then AND it.
Toggling a bit
The XOR operator (
^) can be used to toggle a bit.
number ^= 1UL << n;
That will toggle the
nth bit of
Checking a bit
You didn't ask for this, but I might as well add it.
To check a bit, shift the number n to the right, then bitwise AND it:
bit = (number >> n) & 1U;
That will put the value of the
nth bit of
number into the variable
Changing the nth bit to x
nth bit to either
0 can be achieved with the following on a 2's complement C++ implementation:
number ^= (-x ^ number) & (1UL << n);
n will be set if
1, and cleared if
x has some other value, you get garbage.
x = !!x will booleanize it to 0 or 1.
To make this independent of 2's complement negation behaviour (where
-1 has all bits set, unlike on a 1's complement or sign/magnitude C++ implementation), use unsigned negation.
number ^= (-(unsigned long)x ^ number) & (1UL << n);
unsigned long newbit = !!x; // Also booleanize to force 0 or 1
number ^= (-newbit ^ number) & (1UL << n);
It's generally a good idea to use unsigned types for portable bit manipulation.
number = (number & ~(1UL << n)) | (x << n);
(number & ~(1UL << n)) will clear the
nth bit and
(x << n) will set the
nth bit to
It's also generally a good idea to not to copy/paste code in general and so many people use preprocessor macros (like the community wiki answer further down) or some sort of encapsulation.
This is known as the 'Hamming Weight', 'popcount' or 'sideways addition'.
Some CPUs have a single built-in instruction to do it and others have parallel instructions which act on bit vectors. Instructions like x86's
popcnt (on CPUs where it's supported) will almost certainly be fastest for a single integer. Some other architectures may have a slow instruction implemented with a microcoded loop that tests a bit per cycle (citation needed - hardware popcount is normally fast if it exists at all.).
The 'best' algorithm really depends on which CPU you are on and what your usage pattern is.
Your compiler may know how to do something that's good for the specific CPU you're compiling for, e.g. C++20
std::popcount(), or C++
std::bitset<32>::count(), as a portable way to access builtin / intrinsic functions (see another answer on this question). But your compiler's choice of fallback for target CPUs that don't have hardware popcnt might not be optimal for your use-case. Or your language (e.g. C) might not expose any portable function that could use a CPU-specific popcount when there is one.
Portable algorithms that don't need (or benefit from) any HW support
A pre-populated table lookup method can be very fast if your CPU has a large cache and you are doing lots of these operations in a tight loop. However it can suffer because of the expense of a 'cache miss', where the CPU has to fetch some of the table from main memory. (Look up each byte separately to keep the table small.) If you want popcount for a contiguous range of numbers, only the low byte is changing for groups of 256 numbers, making this very good.
If you know that your bytes will be mostly 0's or mostly 1's then there are efficient algorithms for these scenarios, e.g. clearing the lowest set with a bithack in a loop until it becomes zero.
I believe a very good general purpose algorithm is the following, known as 'parallel' or 'variable-precision SWAR algorithm'. I have expressed this in a C-like pseudo language, you may need to adjust it to work for a particular language (e.g. using uint32_t for C++ and >>> in Java):
GCC10 and clang 10.0 can recognize this pattern / idiom and compile it to a hardware popcnt or equivalent instruction when available, giving you the best of both worlds. (https://godbolt.org/z/qGdh1dvKK)
int numberOfSetBits(uint32_t i)
// Java: use int, and use >>> instead of >>. Or use Integer.bitCount()
// C or C++: use uint32_t
i = i - ((i >> 1) & 0x55555555); // add pairs of bits
i = (i & 0x33333333) + ((i >> 2) & 0x33333333); // quads
i = (i + (i >> 4)) & 0x0F0F0F0F; // groups of 8
return (i * 0x01010101) >> 24; // horizontal sum of bytes
|0 for performance: change the first line to
i = (i|0) - ((i >> 1) & 0x55555555);
This has the best worst-case behaviour of any of the algorithms discussed, so will efficiently deal with any usage pattern or values you throw at it. (Its performance is not data-dependent on normal CPUs where all integer operations including multiply are constant-time. It doesn't get any faster with "simple" inputs, but it's still pretty decent.)
How this SWAR bithack works:
i = i - ((i >> 1) & 0x55555555);
The first step is an optimized version of masking to isolate the odd / even bits, shifting to line them up, and adding. This effectively does 16 separate additions in 2-bit accumulators (SWAR = SIMD Within A Register). Like
(i & 0x55555555) + ((i>>1) & 0x55555555).
The next step takes the odd/even eight of those 16x 2-bit accumulators and adds again, producing 8x 4-bit sums. The
i - ... optimization isn't possible this time so it does just mask before / after shifting. Using the same
0x33... constant both times instead of
0xccc... before shifting is a good thing when compiling for ISAs that need to construct 32-bit constants in registers separately.
The final shift-and-add step of
(i + (i >> 4)) & 0x0F0F0F0F widens to 4x 8-bit accumulators. It masks after adding instead of before, because the maximum value in any 4-bit accumulator is
4, if all 4 bits of the corresponding input bits were set. 4+4 = 8 which still fits in 4 bits, so carry between nibble elements is impossible in
i + (i >> 4).
So far this is just fairly normal SIMD using SWAR techniques with a few clever optimizations. Continuing on with the same pattern for 2 more steps can widen to 2x 16-bit then 1x 32-bit counts. But there is a more efficient way on machines with fast hardware multiply:
Once we have few enough "elements", a multiply with a magic constant can sum all the elements into the top element. In this case byte elements. Multiply is done by left-shifting and adding, so a multiply of
x * 0x01010101 results in
x + (x<<8) + (x<<16) + (x<<24). Our 8-bit elements are wide enough (and holding small enough counts) that this doesn't produce carry into that top 8 bits.
A 64-bit version of this can do 8x 8-bit elements in a 64-bit integer with a 0x0101010101010101 multiplier, and extract the high byte with
>>56. So it doesn't take any extra steps, just wider constants. This is what GCC uses for
__builtin_popcountll on x86 systems when the hardware
popcnt instruction isn't enabled. If you can use builtins or intrinsics for this, do so to give the compiler a chance to do target-specific optimizations.
With full SIMD for wider vectors (e.g. counting a whole array)
This bitwise-SWAR algorithm could parallelize to be done in multiple vector elements at once, instead of in a single integer register, for a speedup on CPUs with SIMD but no usable popcount instruction. (e.g. x86-64 code that has to run on any CPU, not just Nehalem or later.)
However, the best way to use vector instructions for popcount is usually by using a variable-shuffle to do a table-lookup for 4 bits at a time of each byte in parallel. (The 4 bits index a 16 entry table held in a vector register).
On Intel CPUs, the hardware 64bit popcnt instruction can outperform an SSSE3
PSHUFB bit-parallel implementation by about a factor of 2, but only if your compiler gets it just right. Otherwise SSE can come out significantly ahead. Newer compiler versions are aware of the popcnt false dependency problem on Intel.
(This assumes that the top halfword of r1 is empty, like your reference code did.) Depending on the circumstances, you might be able to omit the final mov here by merging it with some later code.