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.
Apart from the apparent difference of
- having to declare the value at the time of a definition for a
const
VS readonly
values can be computed dynamically but need to be assigned before the constructor exits.. after that it is frozen.
const
's are implicitly static
. You use a ClassName.ConstantName
notation to access them.
There is a subtle difference. Consider a class defined in AssemblyA
.
public class Const_V_Readonly
{
public const int I_CONST_VALUE = 2;
public readonly int I_RO_VALUE;
public Const_V_Readonly()
{
I_RO_VALUE = 3;
}
}
AssemblyB
references AssemblyA
and uses these values in code. When this is compiled:
- in the case of the
const
value, it is like a find-replace. The value 2 is 'baked into' the AssemblyB
's IL. This means that if tomorrow I update I_CONST_VALUE
to 20, AssemblyB
would still have 2 till I recompile it.
- in the case of the
readonly
value, it is like a ref
to a memory location. The value is not baked into AssemblyB
's IL. This means that if the memory location is updated, AssemblyB
gets the new value without recompilation. So if I_RO_VALUE
is updated to 30, you only need to build AssemblyA
and all clients do not need to be recompiled.
So if you are confident that the value of the constant won't change, use a const
.
public const int CM_IN_A_METER = 100;
But if you have a constant that may change (e.g. w.r.t. precision).. or when in doubt, use a readonly
.
public readonly float PI = 3.14;
Update: Aku needs to get a mention because he pointed this out first. Also I need to plug where I learned this: Effective C# - Bill Wagner
Best Answer
Without the definition of
SceneTrackerData
, it's hard to say, but likely that function (SceneTrackerData::getRect
) is not marked as const.That is, what is (guessing):
Should be: