An unexpected output

[carter2000]

#include <iostream> using namespace std; class CTestA { public: virtual void FuncA() {} private: int m_iValue1; char m_cValue1; int m_iValue2; char m_cValue2; double m_dValue1; }; class CTestB : public CTestA { public: virtual void FuncB() {} private: }; void main() { cout << sizeof(CTestB) << endl; system("pause"); } Why the output is '32', while my expected is '28'.

[N a v a n e e t h]

carter2000 wrote:

Why the output is '32', while my expected is '28'.

How are you calculating 28? By adding size of individual members? If yes, see what MSDN is saying. When the sizeof operator is applied to a class, struct, or union type, the result is the number of bytes in an object of that type, plus any padding added to align members on word boundaries. The result does not necessarily correspond to the size calculated by adding the storage requirements of the individual members.

Navaneeth How to use google | Ask smart questions

[Stuart Dootson]

Structure sizes don't work that way - there may be (will be in this case) be padding between items of different types in a structure, dependent on the type's default alignment. Have a look at this Wikipedia page[^] that explains the concept. However, there is a way around the default alignments, using #pragma pack to alter structure packing. So, if you put the following line before your structure definitions, you'll find you get the output you expect:

#pragma pack(1)

Java, Basic, who cares - it's all a bunch of tree-hugging hippy cr*p

[carter2000]

Thanks for your answer.

[carter2000]

Thanks. I add "#pragma pack(4)", and it works as I wish.

[Stuart Dootson]

Not #pragma pack(1)?

Java, Basic, who cares - it's all a bunch of tree-hugging hippy cr*p

[carter2000]

Yes, not #pragma pack(1). I put it in my program, and the output was "22". :)