学步园

There are many factors that decide the size of an object of a class in C++.
These factors are:

1. Size of all non-static data members
2. Order of data members
3. Byte alignment or byte padding
4. Size of its immediate base class
5. The existence of virtual function(s) (Dynamic polymorphism using virtual functions).
6. Compiler being used
7. Mode of inheritance (virtual inheritance)

Size of all non-static data members

Only non-static data members will be counted for calculating sizeof
class/object.

class A { 
private: 
        float iMem1; 
        const int iMem2; 
        static int iMem3; 
        char iMem4; 
}; 

For an object of class A, the size will be the size of float iMem1 + size of
int iMem2 + size of char iMem3. Static members are really not part of the
class object. They won't be included in object's layout.

<2>Order of data members

The order in which one specifies data members also alters the size of the
class.

class C { 
        char c; 
        int int1; 
        int int2; 
        int i; 
        long l; 
        short s; 
}; 

The size of this class is 24 bytes. Even though char c will consume
only 1 byte, 4 bytes will be allocated for it, and the remaining 3 bytes will
be wasted (holes). This is because the next member is an int, which takes 4
bytes. If we don't go to the next (4th) byte for storing this integer member, the
memory access/modify cycle for this integer will be 2 read cycles. So the
compiler will do this for us, unless we specify some byte padding/packing.

If I re-write the above class in different order, keeping all my data members like below:

class C { 
        int int1; 
        int int2; 
        int i; 
        long l; 
        short s; 
        char c; 
}; 

Now the size of this class is 20 bytes.

In this case, it is storing c, the char, in one of the slots in the hole in
the extra four bytes.

Byte alignment or byte padding

As mentioned above, if we specify 1 byte alignment, the size of the class above
(class C) will be 19 in both cases.

Size of its immediate base class

The size of a class also includes size of its immediate base class.

Lets take an example:

Class B { 
... 
        int iMem1; 
        int iMem2; 
} 

Class D: public B { 
... 
        int iMem; 
} 

In this case, sizeof(D) is will also include the size of B. So it will be 12
bytes.

The existence of virtual function(s)

Existence of virtual function(s) will add 4 bytes of virtual table pointer in
the class, which will be added to size of class. Again, in this case, if the
base class of the class already has virtual function(s) either directly or
through its base class, then this additional virtual function won't add
anything to the size of the class.

Virtual table pointer will be common across the class hierarchy.

That is

class Base { 
public: 
... 
        virtual void SomeFunction(...); 
private: 
        int iAMem 
}; 

class Derived : public Base { 
... 
        virtual void SomeOtherFunction(...); 
private: 
        int iBMem 
}; 

In the example above, sizeof(Base) will be 8 bytes--that is sizeof(int iAMem) +
sizeof(vptr). sizeof(Derived) will be 12 bytes, that is sizeof(int iBMem) +
sizeof(Derived). Notice that the existence of virtual functions in class
Derived won't add anything more. Now Derived will set the vptr to its own
virtual function table.

Compiler being used

In some scenarios, the size of a class object can be compiler specific.

Lets take one example:

class BaseClass { 
        int a; 
        char c; 
}; 

class DerivedClass : public BaseClass { 
        char d; 
        int i; 
}; 

If compiled with the Microsoft C++ compiler, the size of DerivedClass is 16 bytes. If compiled with
gcc (either c++ or g++), size of DerivedClass is 12 bytes.

The reason for sizeof(DerivedClass) being 16 bytes in MC++ is that it starts each
class with a 4 byte aligned address so that accessing the member of that class
will be easy (again, the memory read/write cycle).

Mode of inheritance (virtual inheritance)

In C++, sometimes we have to use virtual inheritance for some reasons. (One
classic example is the implementation of final class in C++.) When we use
virtual inheritance, there will be the overhead of 4 bytes for a virtual base
class pointer in that class.

class ABase{ 
        int iMem; 
}; 

class BBase : public virtual ABase { 
        int iMem; 
}; 

class CBase : public virtual ABase { 
        int iMem; 
}; 

class ABCDerived : public BBase, public CBase { 
        int iMem; 
}; 

And if you check the size of these classes, it will be:

• Size of ABase : 4
• Size of BBase : 12
• Size of CBase : 12
• Size of ABCDerived : 24

Because BBase and CBase are dervied from ABase virtually, they will also have an
virtual base pointer. So, 4 bytes will be added to the size of the class (BBase
and CBase). That is sizeof ABase + size of int + sizeof Virtual Base pointer.

Size of ABCDerived will be 24 (not 28 = sizeof (BBase + CBase + int member))
because it will maintain only one Virtual Base pointer (Same way of
maintaining virtual table pointer).