c++  program  toggle all bits in c
How do you set, clear, and toggle a single bit? (18)
How do you set, clear, and toggle a bit in C/C++?
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
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 mostsignificantbits.
// 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.
number = (1ull << x);
number = ((uintmax_t)1 << x);
Or cast  which makes for coding/review/maintenance issues keeping the cast correct and uptodate.
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
.
number = (number*0 + 1) << x;
As with most bit manipulations, best to work with unsigned types rather than signed ones
Setting a bit
Use the bitwise OR operator (
) to set a bit.
number = 1UL << n;
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 n1
, 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.
number &= ~(1UL << n);
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.
number ^= 1UL << n;
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:
bit = (number >> n) & 1U;
That will put the value of the n
th bit of number
into the variable bit
.
Changing the nth bit to x
Setting the n
th bit to either 1
or 0
can be achieved with the following on a 2's complement C++ implementation:
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.
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);
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.
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.
From snipc.zip's bitops.h:
/*
** 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:\BINK\tst.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 ]
A C++11 templated version (put in a header):
namespace bit {
template <typename T1, typename T2> inline void set (T1 &variable, T2 bit) {variable = ((T1)1 << bit);}
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 = bits;}
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;}
}
Expanding on the bitset
answer:
#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;
}
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):
#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.
I use macros defined in a header file to handle bit set and clear:
/* a=target variable, b=bit number to act upon 0n */
#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))
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] = TQuickByteMask[n]; // Set bit.
}
/** 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 bigendian processors, ah, I feel a flame war coming on...).
It is sometimes worth using an enum
to name the bits:
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.
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; // onebit flag field
unsigned int Mode:3; // threebit mode field
unsigned int StatusCode:4; // fourbit 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 implementationdependent. Also, verify how your compiler handlers fields crossing byte boundaries.
You can then read, write, test the individual values as before.
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 3bit field (actually, it's three 1bit 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 fixedsize bit fields. Otherwise you have to resort to the bittwiddling techniques described in previous posts.
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)
{
bitfield = (bitfield  (1 << n)) & (~( (1 << n) ^ (value << n) ));
}
Or
void chang_n_bit(int n, int value)
{
bitfield = (bitfield  (1 << n)) & ((value << n)  ((~0) ^ (1 << n)));
}
Or
void chang_n_bit(int n, int value)
{
if(value)
bitfield = 1 << n;
else
bitfield &= ~0 ^ (1 << n);
}
char get_n_bit(int n)
{
return (bitfield & (1 << n)) ? 1 : 0;
}
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.
Use this:
int ToggleNthBit ( unsigned char n, int num )
{
if(num & (1 << n))
num &= ~(1 << n);
else
num = (1 << n);
return num;
}
Using the Standard C++ Library: std::bitset<N>
.
Or the Boost version: 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[04] valid
std::cout << x << std::endl;
}
[Alpha:] > ./a.out
00010
The Boost version allows a runtime sized bitset compared with a standard library compiletime sized bitset.
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 UNprintable 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 6digit 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.
int set_nth_bit(int num, int n){
return (num  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);
}