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
How are virtual functions implemented at a deep level?
From "Virtual Functions in C++":
Can the vtable be modified or even directly accessed at runtime?
Universally, I believe the answer is "no". You could do some memory mangling to find the vtable but you still wouldn't know what the function signature looks like to call it. Anything that you would want to achieve with this ability (that the language supports) should be possible without access to the vtable directly or modifying it at runtime. Also note, the C++ language spec does not specify that vtables are required - however that is how most compilers implement virtual functions.
Does the vtable exist for all objects, or only those that have at least one virtual function?
I believe the answer here is "it depends on the implementation" since the spec doesn't require vtables in the first place. However, in practice, I believe all modern compilers only create a vtable if a class has at least 1 virtual function. There is a space overhead associated with the vtable and a time overhead associated with calling a virtual function vs a non-virtual function.
Do abstract classes simply have a NULL for the function pointer of at least one entry?
The answer is it is unspecified by the language spec so it depends on the implementation. Calling the pure virtual function results in undefined behavior if it is not defined (which it usually isn't) (ISO/IEC 14882:2003 10.4-2). In practice it does allocate a slot in the vtable for the function but does not assign an address to it. This leaves the vtable incomplete which requires the derived classes to implement the function and complete the vtable. Some implementations do simply place a NULL pointer in the vtable entry; other implementations place a pointer to a dummy method that does something similar to an assertion.
Note that an abstract class can define an implementation for a pure virtual function, but that function can only be called with a qualified-id syntax (ie., fully specifying the class in the method name, similar to calling a base class method from a derived class). This is done to provide an easy to use default implementation, while still requiring that a derived class provide an override.
Does having a single virtual function slow down the whole class or only the call to the function that is virtual?
This is getting to the edge of my knowledge, so someone please help me out here if I'm wrong!
I believe that only the functions that are virtual in the class experience the time performance hit related to calling a virtual function vs. a non-virtual function. The space overhead for the class is there either way. Note that if there is a vtable, there is only 1 per class, not one per object.
Does the speed get affected if the virtual function is actually overridden or not, or does this have no effect so long as it is virtual?
I don't believe the execution time of a virtual function that is overridden decreases compared to calling the base virtual function. However, there is an additional space overhead for the class associated with defining another vtable for the derived class vs the base class.
Additional Resources:
http://www.codersource.net/published/view/325/virtual_functions_in.aspx (via way back machine)
http://en.wikipedia.org/wiki/Virtual_table
http://www.codesourcery.com/public/cxx-abi/abi.html#vtable