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 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.
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.
or
number = (number & ~(1UL << n)) | (x << n);
(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.
I see you're using unsigned integers. By definition, in C (I don't know about C++), unsigned arithmetic does not overflow ... so, at least for C, your point is moot :)
With signed integers, once there has been overflow, undefined behaviour (UB) has occurred and your program can do anything (for example: render tests inconclusive).
#include <limits.h>
int a = <something>;
int x = <something>;
a += x; /* UB */
if (a < 0) { /* Unreliable test */
/* ... */
}
To create a conforming program, you need to test for overflow before generating said overflow. The method can be used with unsigned integers too:
// For addition
#include <limits.h>
int a = <something>;
int x = <something>;
if ((x > 0) && (a > INT_MAX - x)) /* `a + x` would overflow */;
if ((x < 0) && (a < INT_MIN - x)) /* `a + x` would underflow */;
// For subtraction
#include <limits.h>
int a = <something>;
int x = <something>;
if ((x < 0) && (a > INT_MAX + x)) /* `a - x` would overflow */;
if ((x > 0) && (a < INT_MIN + x)) /* `a - x` would underflow */;
// For multiplication
#include <limits.h>
int a = <something>;
int x = <something>;
// There may be a need to check for -1 for two's complement machines.
// If one number is -1 and another is INT_MIN, multiplying them we get abs(INT_MIN) which is 1 higher than INT_MAX
if ((a == -1) && (x == INT_MIN)) /* `a * x` can overflow */
if ((x == -1) && (a == INT_MIN)) /* `a * x` (or `a / x`) can overflow */
// general case
if (a > INT_MAX / x) /* `a * x` would overflow */;
if ((a < INT_MIN / x)) /* `a * x` would underflow */;
For division (except for the INT_MIN
and -1
special case), there isn't any possibility of going over INT_MIN
or INT_MAX
.
Best Answer
If you are on a recent-ish x86 or x86-64 platform (and you probably are), use the
bsr
instruction which will return the position of the highest set bit in an unsigned integer. It turns out that this is exactly the same as log2(). Here is a short C or C++ function that invokesbsr
using inline ASM: