I have a small doubt in Virtual Table, whenever compiler encounters the virtual functions in a class, it creates Vtable and places virtual functions address over there. It happens similarly for other class which inherits. Does it create a new pointer in each class which points to each Vtable? If not how does it access the Virtual function when the new instance of derived class is created and assigned to Base PTR?
C++ – How Vtable of Virtual functions work
cvirtualvtable
Related Solutions
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 use this to split string by a delimiter. The first puts the results in a pre-constructed vector, the second returns a new vector.
#include <string>
#include <sstream>
#include <vector>
#include <iterator>
template <typename Out>
void split(const std::string &s, char delim, Out result) {
std::istringstream iss(s);
std::string item;
while (std::getline(iss, item, delim)) {
*result++ = item;
}
}
std::vector<std::string> split(const std::string &s, char delim) {
std::vector<std::string> elems;
split(s, delim, std::back_inserter(elems));
return elems;
}
Note that this solution does not skip empty tokens, so the following will find 4 items, one of which is empty:
std::vector<std::string> x = split("one:two::three", ':');
Best Answer
Each time you create a class that contains virtual functions, or you derive from a class that contains virtual functions, the compiler creates a unique VTABLE for that class.
If you don’t override a function that was declared virtual in the base class, the compiler uses the address of the base-class version in the derived class.
Then it places the VPTR into the class. There is only one VPTR for each object when using simple inheritance . The VPTR must be initialized to point to the starting address of the appropriate VTABLE. (This happens in the constructor.) Once the VPTR is initialized to the proper VTABLE, the object in effect “knows” what type it is. But this self-knowledge is worthless unless it is used at the point a virtual function is called. When you call a virtual function through a base class address (the situation when the compiler doesn’t have all the information necessary to perform early binding), something special happens. Instead of performing a typical function call, which is simply an assembly-language CALL to a particular address, the compiler generates different code to perform the function call.