How do you set, clear, and toggle a bit?

# C++ – How to set, clear, and toggle a single bit

bit-manipulationbitwise-operatorsc

#### Related Solutions

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
}
```

For JavaScript: coerce to integer with `|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.)

References:

- https://graphics.stanford.edu/~seander/bithacks.html
- https://en.wikipedia.org/wiki/Hamming_weight
- http://gurmeet.net/puzzles/fast-bit-counting-routines/
- http://aggregate.ee.engr.uky.edu/MAGIC/#Population%20Count%20(Ones%20Count)

### 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.

- https://github.com/WojciechMula/sse-popcount state-of-the-art x86 SIMD popcount for SSSE3, AVX2, AVX512BW, AVX512VBMI, or AVX512 VPOPCNT. Using Harley-Seal across vectors to defer popcount within an element. (Also ARM NEON)
- Counting 1 bits (population count) on large data using AVX-512 or AVX-2
- related: https://github.com/mklarqvist/positional-popcount - separate counts for each bit-position of multiple 8, 16, 32, or 64-bit integers. (Again, x86 SIMD including AVX-512 which is really good at this, with
`vpternlogd`

making Harley-Seal*very*good.)

The bit shifting operators do exactly what their name implies. They shift bits. Here's a brief (or not-so-brief) introduction to the different shift operators.

## The Operators

`>>`

is the arithmetic (or signed) right shift operator.`>>>`

is the logical (or unsigned) right shift operator.`<<`

is the left shift operator, and meets the needs of both logical and arithmetic shifts.

All of these operators can be applied to integer values (`int`

, `long`

, possibly `short`

and `byte`

or `char`

). In some languages, applying the shift operators to any datatype smaller than `int`

automatically resizes the operand to be an `int`

.

Note that `<<<`

is not an operator, because it would be redundant.

Also note that **C and C++ do not distinguish between the right shift operators**. They provide only the `>>`

operator, and the right-shifting behavior is implementation defined for signed types. The rest of the answer uses the C# / Java operators.

(In all mainstream C and C++ implementations including GCC and Clang/LLVM, `>>`

on signed types is arithmetic. Some code assumes this, but it isn't something the standard guarantees. It's not *undefined*, though; the standard requires implementations to define it one way or another. However, left shifts of negative signed numbers *is* undefined behaviour (signed integer overflow). So unless you need arithmetic right shift, it's usually a good idea to do your bit-shifting with unsigned types.)

## Left shift (<<)

Integers are stored, in memory, as a series of bits. For example, the number 6 stored as a 32-bit `int`

would be:

```
00000000 00000000 00000000 00000110
```

Shifting this bit pattern to the left one position (`6 << 1`

) would result in the number 12:

```
00000000 00000000 00000000 00001100
```

As you can see, the digits have shifted to the left by one position, and the last digit on the right is filled with a zero. You might also note that shifting left is equivalent to multiplication by powers of 2. So `6 << 1`

is equivalent to `6 * 2`

, and `6 << 3`

is equivalent to `6 * 8`

. A good optimizing compiler will replace multiplications with shifts when possible.

### Non-circular shifting

Please note that these are *not* circular shifts. Shifting this value to the left by one position (`3,758,096,384 << 1`

):

```
11100000 00000000 00000000 00000000
```

results in 3,221,225,472:

```
11000000 00000000 00000000 00000000
```

The digit that gets shifted "off the end" is lost. It does not wrap around.

## Logical right shift (>>>)

A logical right shift is the converse to the left shift. Rather than moving bits to the left, they simply move to the right. For example, shifting the number 12:

```
00000000 00000000 00000000 00001100
```

to the right by one position (`12 >>> 1`

) will get back our original 6:

```
00000000 00000000 00000000 00000110
```

So we see that shifting to the right is equivalent to division by powers of 2.

### Lost bits are gone

However, a shift cannot reclaim "lost" bits. For example, if we shift this pattern:

```
00111000 00000000 00000000 00000110
```

to the left 4 positions (`939,524,102 << 4`

), we get 2,147,483,744:

```
10000000 00000000 00000000 01100000
```

and then shifting back (`(939,524,102 << 4) >>> 4`

) we get 134,217,734:

```
00001000 00000000 00000000 00000110
```

We cannot get back our original value once we have lost bits.

# Arithmetic right shift (>>)

The arithmetic right shift is exactly like the logical right shift, except instead of padding with zero, it pads with the most significant bit. This is because the most significant bit is the *sign* bit, or the bit that distinguishes positive and negative numbers. By padding with the most significant bit, the arithmetic right shift is sign-preserving.

For example, if we interpret this bit pattern as a negative number:

```
10000000 00000000 00000000 01100000
```

we have the number -2,147,483,552. Shifting this to the right 4 positions with the arithmetic shift (-2,147,483,552 >> 4) would give us:

```
11111000 00000000 00000000 00000110
```

or the number -134,217,722.

So we see that we have preserved the sign of our negative numbers by using the arithmetic right shift, rather than the logical right shift. And once again, we see that we are performing division by powers of 2.

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## Best Answer

## Setting a bit

Use the bitwise OR operator (

`|`

) to set a bit.That will set the

`n`

th bit of`number`

.`n`

should be zero, if you want to set the`1`

st bit and so on upto`n-1`

, if you want to set the`n`

th bit.Use

`1ULL`

if`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.That will clear the

`n`

th 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.That will toggle the

`n`

th bit of`number`

.## 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:

That will put the value of the

`n`

th bit of`number`

into the variable`bit`

.## Changing the

nth bit toxSetting the

`n`

th bit to either`1`

or`0`

can be achieved with the following on a 2's complement C++ implementation:Bit

`n`

will be set if`x`

is`1`

, and cleared if`x`

is`0`

. 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.or

It's generally a good idea to use unsigned types for portable bit manipulation.

or

`(number & ~(1UL << n))`

will clear the`n`

th bit and`(x << n)`

will set the`n`

th bit to`x`

.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.