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

How do you set, clear, and toggle a bit in C/C++?

26 Answers
26

Setting a bit

Use the bitwise OR operator (|) to set a bit.

|

number |= 1UL << n;


That will set the nth bit of number. n should be zero, if you want to set the 1st bit and so on upto n-1, if you want to set the nth bit.

n

number

n

1

n-1

n

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.

1ULL

number

unsigned long

1UL << n

1UL << n

long

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.

n

number

~

Toggling a bit

The XOR operator (^) can be used to toggle a bit.

^

number ^= 1UL << n;


That will toggle the nth bit of number.

n

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:

bit = (number >> n) & 1U;


That will put the value of the nth bit of number into the variable bit.

n

number

bit

Changing the nth bit to x

Setting the nth bit to either 1 or 0 can be achieved with the following on a 2's complement C++ implementation:

n

1

0

number ^= (-x ^ number) & (1UL << n);


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.

n

x

1

x

0

x

x = !!x

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.

-1

number ^= (-(unsigned long)x ^ number) & (1UL << n);


or

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.

or

number = (number & ~(1UL << n)) | (x << n);


(number & ~(1UL << n)) will clear the nth bit and (x << n) will set the nth bit to x.

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

n

(x << n)

n

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.

bit = (number >> x) & 1

1

int

1U

number = number & ~(1 << n) | (x << n);

Using the Standard C++ Library: std::bitset<N>.

std::bitset<N>

Or the Boost version: boost::dynamic_bitset.

boost::dynamic_bitset

There is no need to roll your own:

#include <bitset>
#include <iostream>

int main()

std::bitset<5> x;

x[1] = 1;
x[2] = 0;
// Note x[0-4] valid

std::cout << x << std::endl;



[Alpha:] > ./a.out
00010


The Boost version allows a runtime sized bitset compared with a standard library compile-time sized bitset.

The other option is to use bit fields:

struct bits
unsigned int a:1;
unsigned int b:1;
unsigned int c:1;
;

struct bits mybits;


defines a 3-bit field (actually, it's three 1-bit felds). Bit operations now become a bit (haha) simpler:

To set or clear a bit:

mybits.b = 1;
mybits.c = 0;


To toggle a bit:

mybits.a = !mybits.a;
mybits.b = ~mybits.b;
mybits.c ^= 1; /* all work */


Checking a bit:

if (mybits.c) //if mybits.c is non zero the next line below will execute


This only works with fixed-size bit fields. Otherwise you have to resort to the bit-twiddling techniques described in previous posts.

I use macros defined in a header file to handle bit set and clear:

/* a=target variable, b=bit number to act upon 0-n */
#define BIT_SET(a,b) ((a) |= (1ULL<<(b)))
#define BIT_CLEAR(a,b) ((a) &= ~(1ULL<<(b)))
#define BIT_FLIP(a,b) ((a) ^= (1ULL<<(b)))
#define BIT_CHECK(a,b) (!!((a) & (1ULL<<(b)))) // '!!' to make sure this returns 0 or 1

/* x=target variable, y=mask */
#define BITMASK_SET(x,y) ((x) |= (y))
#define BITMASK_CLEAR(x,y) ((x) &= (~(y)))
#define BITMASK_FLIP(x,y) ((x) ^= (y))
#define BITMASK_CHECK_ALL(x,y) (((x) & (y)) == (y)) // warning: evaluates y twice
#define BITMASK_CHECK_ANY(x,y) ((x) & (y))


BITMASK_CHECK(x,y) ((x) & (y))

((x) & (y)) == (y)

5

3

1

(uintmax_t)1

long

1ULL

(uintmax_t)

It is sometimes worth using an enum to name the bits:

enum

enum ThingFlags =
ThingMask = 0x0000,
ThingFlag0 = 1 << 0,
ThingFlag1 = 1 << 1,
ThingError = 1 << 8,



Then use the names later on. I.e. write

thingstate |= ThingFlag1;
thingstate &= ~ThingFlag0;
if (thing & ThingError) ...


to set, clear and test. This way you hide the magic numbers from the rest of your code.

Other than that I endorse Jeremy's solution.

clearbits()

&= ~

enum

const short

enum ThingFlags

ThingError|ThingFlag1

int

thingstate = ThingFlag1 >> 1

thingstate = (ThingFlag1 >> x) << y

enum My16Bits: unsigned short ... ;

/*
** Bit set, clear, and test operations
**
** public domain snippet by Bob Stout
*/

typedef enum ERROR = -1, FALSE, TRUE LOGICAL;

#define BOOL(x) (!(!(x)))

#define BitSet(arg,posn) ((arg) | (1L << (posn)))
#define BitClr(arg,posn) ((arg) & ~(1L << (posn)))
#define BitTst(arg,posn) BOOL((arg) & (1L << (posn)))
#define BitFlp(arg,posn) ((arg) ^ (1L << (posn)))


OK, let's analyze things...

The common expression that you seem to be having problems with in all of these is "(1L << (posn))". All this does is create a mask with a single bit on
and which will work with any integer type. The "posn" argument specifies the
position where you want the bit. If posn==0, then this expression will
evaluate to:

0000 0000 0000 0000 0000 0000 0000 0001 binary.


If posn==8, it will evaluate to

0000 0000 0000 0000 0000 0001 0000 0000 binary.


In other words, it simply creates a field of 0's with a 1 at the specified
position. The only tricky part is in the BitClr() macro where we need to set
a single 0 bit in a field of 1's. This is accomplished by using the 1's
complement of the same expression as denoted by the tilde (~) operator.

Once the mask is created it's applied to the argument just as you suggest,
by use of the bitwise and (&), or (|), and xor (^) operators. Since the mask
is of type long, the macros will work just as well on char's, short's, int's,
or long's.

The bottom line is that this is a general solution to an entire class of
problems. It is, of course, possible and even appropriate to rewrite the
equivalent of any of these macros with explicit mask values every time you
need one, but why do it? Remember, the macro substitution occurs in the
preprocessor and so the generated code will reflect the fact that the values
are considered constant by the compiler - i.e. it's just as efficient to use
the generalized macros as to "reinvent the wheel" every time you need to do
bit manipulation.

Unconvinced? Here's some test code - I used Watcom C with full optimization
and without using _cdecl so the resulting disassembly would be as clean as
possible:

----[ TEST.C ]----------------------------------------------------------------

#define BOOL(x) (!(!(x)))

#define BitSet(arg,posn) ((arg) | (1L << (posn)))
#define BitClr(arg,posn) ((arg) & ~(1L << (posn)))
#define BitTst(arg,posn) BOOL((arg) & (1L << (posn)))
#define BitFlp(arg,posn) ((arg) ^ (1L << (posn)))

int bitmanip(int word)

word = BitSet(word, 2);
word = BitSet(word, 7);
word = BitClr(word, 3);
word = BitFlp(word, 9);
return word;



----[ TEST.OUT (disassembled) ]-----------------------------------------------

Module: C:BINKtst.c
Group: 'DGROUP' CONST,CONST2,_DATA,_BSS

Segment: _TEXT BYTE 00000008 bytes
0000 0c 84 bitmanip_ or al,84H ; set bits 2 and 7
0002 80 f4 02 xor ah,02H ; flip bit 9 of EAX (bit 1 of AH)
0005 24 f7 and al,0f7H
0007 c3 ret

No disassembly errors


----[ finis ]-----------------------------------------------------------------

arg

long long

1L

(uintmax_t)1

1ull

Use the bitwise operators: & |

&

|

To set last bit in 000b:

000b

foo = foo | 001b


To check last bit in foo:

foo

if ( foo & 001b ) ....


To clear last bit in foo:

foo

foo = foo & 110b


I used XXXb for clarity. You'll probably be working with HEX representation, depending on the data structure in which you're packing bits.

XXXb

For the beginner I would like to explain a bit more with an example:

Example:

value is 0x55;
bitnum : 3rd.


The & operator is used check the bit:

&

0101 0101
&
0000 1000
___________
0000 0000 (mean 0: False). It will work fine if the third bit is 1 (then the answer will be True)


Toggle or Flip:

0101 0101
^
0000 1000
___________
0101 1101 (Flip the third bit without affecting other bits)


| operator: set the bit

|

0101 0101
|
0000 1000
___________
0101 1101 (set the third bit without affecting other bits)


Here's my favorite bit arithmetic macro, which works for any type of unsigned integer array from unsigned char up to size_t (which is the biggest type that should be efficient to work with):

unsigned char

size_t

#define BITOP(a,b,op)
((a)[(size_t)(b)/(8*sizeof *(a))] op ((size_t)1<<((size_t)(b)%(8*sizeof *(a)))))


To set a bit:

BITOP(array, bit, |=);


To clear a bit:

BITOP(array, bit, &=~);


To toggle a bit:

BITOP(array, bit, ^=);


To test a bit:

if (BITOP(array, bit, &)) ...


etc.

BITOP(array, bit++, |=);

BITCELL(a,b) |= BITMASK(a,b);

a

a

sizeof

(size_t)

%

(unsigned)

(size_t)(b)/(8*sizeof *(a))

b

The bitfield approach has other advantages in the embedded arena. You can define a struct that maps directly onto the bits in a particular hardware register.

struct HwRegister
unsigned int errorFlag:1; // one-bit flag field
unsigned int Mode:3; // three-bit mode field
unsigned int StatusCode:4; // four-bit status code
;

struct HwRegister CR3342_AReg;


You need to be aware of the bit packing order - I think it's MSB first, but this may be implementation-dependent. Also, verify how your compiler handlers fields crossing byte boundaries.

You can then read, write, test the individual values as before.

As this is tagged "embedded" I'll assume you're using a microcontroller. All of the above suggestions are valid & work (read-modify-write, unions, structs, etc.).

However, during a bout of oscilloscope-based debugging I was amazed to find that these methods have a considerable overhead in CPU cycles compared to writing a value directly to the micro's PORTnSET / PORTnCLEAR registers which makes a real difference where there are tight loops / high-frequency ISR's toggling pins.

For those unfamiliar: In my example, the micro has a general pin-state register PORTn which reflects the output pins, so doing PORTn |= BIT_TO_SET results in a read-modify-write to that register. However, the PORTnSET / PORTnCLEAR registers take a '1' to mean "please make this bit 1" (SET) or "please make this bit zero" (CLEAR) and a '0' to mean "leave the pin alone". so, you end up with two port addresses depending whether you're setting or clearing the bit (not always convenient) but a much faster reaction and smaller assembled code.

volatile

More general, for arbitrary sized bitmaps:

#define BITS 8
#define BIT_SET( p, n) (p[(n)/BITS] |= (0x80>>((n)%BITS)))
#define BIT_CLEAR(p, n) (p[(n)/BITS] &= ~(0x80>>((n)%BITS)))
#define BIT_ISSET(p, n) (p[(n)/BITS] & (0x80>>((n)%BITS)))


CHAR_BIT

limits.h

BITS

#define bit_test(x, y) ( ( ((const char*)&(x))[(y)>>3] & 0x80 >> ((y)&0x07)) >> (7-((y)&0x07) ) )


Sample usage:

int main(void)

unsigned char arr[8] = 0x01, 0x23, 0x45, 0x67, 0x89, 0xAB, 0xCD, 0xEF ;

for (int ix = 0; ix < 64; ++ix)
printf("bit %d is %dn", ix, bit_test(arr, ix));

return 0;



Notes:
This is designed to be fast (given its flexibility) and non-branchy. It results in efficient SPARC machine code when compiled Sun Studio 8; I've also tested it using MSVC++ 2008 on amd64. It's possible to make similar macros for setting and clearing bits. The key difference of this solution compared with many others here is that it works for any location in pretty much any type of variable.

This program is to change any data bit from 0 to 1 or 1 to 0:


unsigned int data = 0x000000F0;
int bitpos = 4;
int bitvalue = 1;
unsigned int bit = data;
bit = (bit>>bitpos)&0x00000001;
int invbitvalue = 0x00000001&(~bitvalue);
printf("%xn",bit);

if (bitvalue == 0)

if (bit == 0)
printf("%xn", data);
else

data = (data^(invbitvalue<<bitpos));
printf("%xn", data);


else

if (bit == 1)
printf("elseif %xn", data);
else
(bitvalue<<bitpos));
printf("else %xn", data);





If you're doing a lot of bit twiddling you might want to use masks which will make the whole thing quicker. The following functions are very fast and are still flexible (they allow bit twiddling in bit maps of any size).

const unsigned char TQuickByteMask[8] =

0x01, 0x02, 0x04, 0x08,
0x10, 0x20, 0x40, 0x80,
;


/** Set bit in any sized bit mask.
*
* @return none
*
* @param bit - Bit number.
* @param bitmap - Pointer to bitmap.
*/
void TSetBit( short bit, unsigned char *bitmap)

short n, x;

x = bit / 8; // Index to byte.
n = bit % 8; // Specific bit in byte.

bitmap[x]


/** Reset bit in any sized mask.
*
* @return None
*
* @param bit - Bit number.
* @param bitmap - Pointer to bitmap.
*/
void TResetBit( short bit, unsigned char *bitmap)

short n, x;

x = bit / 8; // Index to byte.
n = bit % 8; // Specific bit in byte.

bitmap[x] &= (~TQuickByteMask[n]); // Reset bit.



/** Toggle bit in any sized bit mask.
*
* @return none
*
* @param bit - Bit number.
* @param bitmap - Pointer to bitmap.
*/
void TToggleBit( short bit, unsigned char *bitmap)

short n, x;

x = bit / 8; // Index to byte.
n = bit % 8; // Specific bit in byte.

bitmap[x] ^= TQuickByteMask[n]; // Toggle bit.



/** Checks specified bit.
*
* @return 1 if bit set else 0.
*
* @param bit - Bit number.
* @param bitmap - Pointer to bitmap.
*/
short TIsBitSet( short bit, const unsigned char *bitmap)

short n, x;

x = bit / 8; // Index to byte.
n = bit % 8; // Specific bit in byte.

// Test bit (logigal AND).
if (bitmap[x] & TQuickByteMask[n])
return 1;

return 0;



/** Checks specified bit.
*
* @return 1 if bit reset else 0.
*
* @param bit - Bit number.
* @param bitmap - Pointer to bitmap.
*/
short TIsBitReset( short bit, const unsigned char *bitmap)

return TIsBitSet(bit, bitmap) ^ 1;



/** Count number of bits set in a bitmap.
*
* @return Number of bits set.
*
* @param bitmap - Pointer to bitmap.
* @param size - Bitmap size (in bits).
*
* @note Not very efficient in terms of execution speed. If you are doing
* some computationally intense stuff you may need a more complex
* implementation which would be faster (especially for big bitmaps).
* See (http://graphics.stanford.edu/~seander/bithacks.html).
*/
int TCountBits( const unsigned char *bitmap, int size)

int i, count = 0;

for (i=0; i<size; i++)
if (TIsBitSet(i, bitmap))
count++;

return count;



Note, to set bit 'n' in a 16 bit integer you do the following:

TSetBit( n, &my_int);


It's up to you to ensure that the bit number is within the range of the bit map that you pass. Note that for little endian processors that bytes, words, dwords, qwords, etc., map correctly to each other in memory (main reason that little endian processors are 'better' than big-endian processors, ah, I feel a flame war coming on...).

Use this:

int ToggleNthBit ( unsigned char n, int num )
= (1 << n);

return num;



Expanding on the bitset answer:

bitset

#include <iostream>
#include <bitset>
#include <string>

using namespace std;
int main()
bitset<8> byte(std::string("10010011");

// Set Bit
byte.set(3); // 10010111

// Clear Bit
byte.reset(2); // 10010101

// Toggle Bit
byte.flip(7); // 00010101

cout << byte << endl;

return 0;



If you want to perform this all operation with C programming in the Linux kernel then I suggest to use standard APIs of the Linux kernel.

See https://www.kernel.org/doc/htmldocs/kernel-api/ch02s03.html

set_bit Atomically set a bit in memory
clear_bit Clears a bit in memory
change_bit Toggle a bit in memory
test_and_set_bit Set a bit and return its old value
test_and_clear_bit Clear a bit and return its old value
test_and_change_bit Change a bit and return its old value
test_bit Determine whether a bit is set


Note: Here the whole operation happens in a single step. So these all are guaranteed to be atomic even on SMP computers and are useful
to keep coherence across processors.

Visual C 2010, and perhaps many other compilers, have direct support for bit operations built in. Surprisingly, this works, even the sizeof() operator works properly.

bool IsGph[256], IsNotGph[256];

// Initialize boolean array to detect printable characters
for(i=0; i<sizeof(IsGph); i++)
IsGph[i] = isgraph((unsigned char)i);



So, to your question, IsGph[i] =1, or IsGph[i] =0 make setting and clearing bools easy.

To find unprintable characters...

// Initialize boolean array to detect UN-printable characters,
// then call function to toggle required bits true, while initializing a 2nd
// boolean array as the complement of the 1st.
for(i=0; i<sizeof(IsGph); i++)
if(IsGph[i])
IsNotGph[i] = 0;
else
IsNotGph[i] = 1;




Note there is nothing "special" about this code. It treats a bit like an integer - which technically, it is. A 1 bit integer that can hold 2 values, and 2 values only.

I once used this approach to find duplicate loan records, where loan_number was the ISAM key, using the 6-digit loan number as an index into the bit array. Savagely fast, and after 8 months, proved that the mainframe system we were getting the data from was in fact malfunctioning. The simplicity of bit arrays makes confidence in their correctness very high - vs a searching approach for example.

bool

int

bool

Use one of the operators as defined here.

To set a bit, used int x = x | 0x?; where ? is the bit position in binary form.

int x = x | 0x?;

?

0x

Here are some macros I use:

SET_FLAG(Status, Flag) ((Status) |= (Flag))
CLEAR_FLAG(Status, Flag) ((Status) &= ~(Flag))
INVALID_FLAGS(ulFlags, ulAllowed) ((ulFlags) & ~(ulAllowed))
TEST_FLAGS(t,ulMask, ulBit) (((t)&(ulMask)) == (ulBit))
IS_FLAG_SET(t,ulMask) TEST_FLAGS(t,ulMask,ulMask)
IS_FLAG_CLEAR(t,ulMask) TEST_FLAGS(t,ulMask,0)


Variable used

int value, pos;


value - Data

pos - position of the bit that we're interested to set, clear or toggle
Set a bit

value = value | 1 << pos;


Clear a bit

value = value & ~(1 << pos);


Toggle a bit

value = value ^ 1 << pos;


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

To address a common coding pitfall when attempting to form the mask:
1 is not always wide enough

1

What problems happen when number is a wider type than 1?
x may be too great for the shift 1 << x leading to undefined behavior (UB). Even if x is not too great, ~ may not flip enough most-significant-bits.

number

1

x

1 << x

x

~

// assume 32 bit int/unsigned
unsigned long long number = foo();

unsigned x = 40;
number |= (1 << x); // UB
number ^= (1 << x); // UB
number &= ~(1 << x); // UB

x = 10;
number &= ~(1 << x); // Wrong mask, not wide enough


To insure 1 is wide enough:

Code could use 1ull or pedantically (uintmax_t)1 and let the compiler optimize.

1ull

(uintmax_t)1

number |= (1ull << x);
number |= ((uintmax_t)1 << x);


Or cast - which makes for coding/review/maintenance issues keeping the cast correct and up-to-date.

number |= (type_of_number)1 << x;


Or gently promote the 1 by forcing a math operation that is as least as wide as the type of number.

1

number

number |= (number*0 + 1) << x;


As with most bit manipulations, best to work with unsigned types rather than signed ones

number |= (type_of_number)1 << x;

number |= (number*0 + 1) << x;

number |= (1ull << x);

int set_nth_bit(int num, int n) 1 << n);


int clear_nth_bit(int num, int n)

return (num & ~( 1 << n));


int toggle_nth_bit(int num, int n)

return num ^ (1 << n);


int check_nth_bit(int num, int n)

return num & (1 << n);



A C++11 templated version (put in a header):

namespace bit
template <typename T1, typename T2> inline void set (T1 &variable, T2 bit) variable
template <typename T1, typename T2> inline void clear(T1 &variable, T2 bit) variable &= ~((T1)1 << bit);
template <typename T1, typename T2> inline void flip (T1 &variable, T2 bit) variable ^= ((T1)1 << bit);
template <typename T1, typename T2> inline bool test (T1 &variable, T2 bit) return variable & ((T1)1 << bit);


namespace bitmask
template <typename T1, typename T2> inline void set (T1 &variable, T2 bits) variable
template <typename T1, typename T2> inline void clear(T1 &variable, T2 bits) variable &= ~bits;
template <typename T1, typename T2> inline void flip (T1 &variable, T2 bits) variable ^= bits;
template <typename T1, typename T2> inline bool test_all(T1 &variable, T2 bits) return ((variable & bits) == bits);
template <typename T1, typename T2> inline bool test_any(T1 &variable, T2 bits) return variable & bits;



;

(variable & bits == bits)

((variable & bits) == bits)

Try one of these functions in the C language to change n bit:

char bitfield;

// Start at 0th position

void chang_n_bit(int n, int value)
(1 << n)) & (~( (1 << n) ^ (value << n) ));



Or

void chang_n_bit(int n, int value)

bitfield = (bitfield


Or

void chang_n_bit(int n, int value)
= 1 << n;
else
bitfield &= ~0 ^ (1 << n);


char get_n_bit(int n)

return (bitfield & (1 << n)) ? 1 : 0;



value << n

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