What is the best way to set a particular bit in a variable in C

Question

Consider a variable unsigned int a; in C.

Now say I want to set any i\'th bit in this variable to \'1\'.

Note that the variable has some value.

Answer 1

Some useful bit manipulation macros

#define BIT_MASK(bit)             (1 << (bit))
#define SET_BIT(value,bit)        ((value) |= BIT_MASK(bit))
#define CLEAR_BIT(value,bit)      ((value) &= ~BIT_MASK(bit))
#define TEST_BIT(value,bit)       (((value) & BIT_MASK(bit)) ? 1 : 0)

Answer 2

You could use a bitwise OR:

a |= (1 << i);

Note that this does not have the same behavior as +, which will carry if there's already a 1 in the bit you're setting.

Answer 3

You should use bitwise OR for this operation.

a |= 1 << i;

Answer 4

Bitwise or it. e.g. a |= (1<<i)

Answer 5

The way I implemented bit flags (to quote straight out of my codebase, you can use it freely for whatever purpose, even commercial):

void SetEnableFlags(int &BitFlags, const int Flags)
{
    BitFlags = (BitFlags|Flags);
}
const int EnableFlags(const int BitFlags, const int Flags)
{
    return (BitFlags|Flags);
}

void SetDisableFlags(int BitFlags, const int Flags)
{
    BitFlags = (BitFlags&(~Flags));
}
const int DisableFlags(const int BitFlags, const int Flags)
{
    return (BitFlags&(~Flags));
}

No bitwise shift operation needed.

You might have to tidy up or alter the code to use the particular variable set you're using, but generally it should work fine.

Answer 6

The most common way to do this is:

a |= (1 << i);

This is only two operations - a shift and an OR. It's hard to see how this might be improved upon.