How counts the operator sizeof(type) the size of a class?

Question

#include <iostream>

class A
{
int i;
char charVec[3];
public:
const int& size() const
{
return sizeof(i) + sizeof(charVec);
}
};

class B
{
int i;
char charVec[5];
public:
const int& size() const
{
return sizeof(i) + sizeof(charVec);
}
};

int main()
{
A a;
B b;
std::cout << "The size of class A with using the operator sizeof: "
<<sizeof(A) << std::endl;
std::cout << "The correct size of class A: " << a.size() << std::endl;
std::cout << "The size of class B with using the operator sizeof: "
<< sizeof(B) << std::endl;
std::cout << "The correct size of class B: " << b.size()<< std::endl;

std::cin.get();
std::cin.get();
return 0;
}

Answer 1

sizeof takes the accumulated object sizes of each member variable.

For example the size of an
int is 4 bytes,
char is 1 byte,
short is 2 bytes,
any pointer is 4 or 8 bytes ( 32 bit and 64 bit OS )
char[10] is 1 * 10 is 10 bytes.

So for class A int + char[3] is 4 + 1 * 3 = 7 bytes.

The compiler usually padds this size to a multiple of 4 or 8 for efficiency reasons, so the size could also be 8.

In gcc you could do
class __attribute((packed)) A{};
then the compiler will not padd it, but don't do that unless you have a good reason to.

Answer 2

I guess (x/4+1)*4, where x is the actual size of the class. But why?

Answer 3

Just wanted to add that if a function is marked virtual then another 4 or 8 bytes ( 32 bit vs 64 bit OS ) is added for the virtual jump table.

Answer 4

Thanks, that was my idea too.