Question

For example, say I have a class Temp:

class Temp
{
    public:
        int function1(int foo) { return 1; }
        void function2(int bar) { foobar = bar; }

    private:
        int foobar;
};

When I create an object of class Temp, how would I calculate how much space it needs, and how is it represented in memory (e.g.| 4 bytes for foobar| 8 bytes for function1 | etc | )

Answer 1

To a first order approximation, the size of an object is the sum of the sizes of its constituent data members. You can be sure it will never be smaller than this.

More precisely, the compiler is entitled to insert padding space between data members to ensure that each data member meets the alignment requirements of the platform. Some platforms are very strict about alignment, while others (x86) are more forgiving, but will perform significantly better with proper alignment. So, even the compiler optimization setting can affect the object size.

Inheritance and virtual functions add an additional complication. As others have said, the member functions of your class themselves do not take up “per object” space, but the existence of virtual functions in that class’s interface generally implies the existence of a virtual table, essentially a lookup table of function pointers used to dynamically resolve the proper function implementation to call at runtime. The virtual table (vtbl) is accessed generally via a pointer stored in each object.

Derived class objects also include all data members of their base classes.

Finally, access specifiers (public, private, protected) grant the compiler certain leeway with packing of data members.

The short answer is that sizeof(myObj) or sizeof(MyClass) will always tell you the proper size of an object, but its result is not always easy to predict.