Size of a Struct does not equal to total size of each member?

There is something called packing alignment which is applied for structure, union, and class members. This alignment can be specified to the compiler by using #pragma instruct.

Now for your compiler I guess the default packing alignment is 4. So once that 4 bytes are exhausted, then the compiler will go for 4 bytes instead of the exact size of the member.

One should be careful while modifying packing alignment as this may effect performance.

Now a struct with members

#pragma pack(4)
struct A
{
   short m_s1;
   int* m_pi;
   short m_s2;
};

void main()
{
     A obj;
     size_t s = sizeof(obj);
}

Here sizeof(obj) is 12. Now we can reduce this to 8 bytes by changing the order of structure members

struct A
{
   short m_s1;
   short m_s2;
   int* m_pi;
};

Thanks for you very quick answer.
Thank to ur answer, I have read about "Struct Aligment" at:
http://msdn.microsoft.com/en-us/library/71kf49f1(v=vs.80).aspx[^]

my answer is closed now.