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.
Best Answer
The phrase "ANSI C" commonly refers to the language defined by the 1989 ISO C standard. The 1990 ISO C standard describes exactly the same language (it adds some ISO-mandated sections). ANSI officially dropped its own 1989 standard and adopted the ISO standard -- and has also adopted the 1999 and 2011 editions of the ISO C standard. So the C standard defined by ANSI is the 2011 ISO C standard.
Because of this ambiguity, I suggest avoiding the phrase "ANSI C" and referring to the year in which the standard was published by ISO: C90, C99, or C11.
There is none.
C90 did not support the and header or the type
long long
. It did not require support for any 64-bit integer type (long
may be 64 bits, but is commonly 32). Because the types didn't exist,printf
provided no format specifiers for them.In principle, you could use a C90 implementation that supports 64-bit
long
(sacrificing portability), or you could implement your own 64-bit type using an array of narrower integers, providing operations as functions.If you could update your question to explain exactly why you want to restrict yourself to a 29-year-old version of the C standard, it might be possible to give a more useful answer. I'd tell you simply to use an implementation that support C99 or C11, but you've indicated that's not a solution (without explaining why).