C++ Object Size

#include <cstdio> class A1 { }; class A2 { public: virtual ~A2() {}; }; class B1 : public A1 { }; class B2 : public A1 { private: char c; }; class B3 : public A1 { private: int i; }; class C1 : public B3 { private: char c; }; class B4 : public A2 { }; class B5 : public A2 { private: char c1; char c2; }; class B6 : public virtual A2 { }; class B7 : public virtual A1 { }; class C2 : public virtual B2 { }; class C3 : public virtual B3 { }; class C4 : public virtual B5 { }; class C5 : virtual public A2 { }; class C6 : virtual public A2 { }; class D1 : public C5, public C6 { }; int main(int argc, char** argv) { printf("size of A1 is = %d/n", sizeof(A1)); printf("size of B1 is = %d/n", sizeof(B1)); printf("size of B2 is = %d/n", sizeof(B2)); printf("size of B3 is = %d/n", sizeof(B3)); printf("size of C1 is = %d/n", sizeof(C1)); printf("size of A2 is = %d/n", sizeof(A2)); printf("size of B4 is = %d/n", sizeof(B4)); printf("size of B5 is = %d/n", sizeof(B5)); printf("size of B6 is = %d/n", sizeof(B6)); printf("size of B7 is = %d/n", sizeof(B7)); printf("size of C2 is = %d/n", sizeof(C2)); printf("size of C3 is = %d/n", sizeof(C3)); printf("size of C4 is = %d/n", sizeof(C4)); printf("size of C5 is = %d/n", sizeof(C5)); printf("size of C6 is = %d/n", sizeof(C5)); printf("size of D1 is = %d/n", sizeof(D1)); return 0; }

 

 

The answer is:

size of A1 is = 1
size of B1 is = 1
size of B2 is = 1
size of B3 is = 4
size of C1 is = 8
size of A2 is = 4
size of B4 is = 4
size of B5 is = 8
size of B6 is = 4
size of B7 is = 4
size of C2 is = 8
size of C3 is = 8
size of C4 is = 12
size of C5 is = 4
size of C6 is = 4
size of D1 is = 8

 

1. 一个空类对象的大小（A1的size）不是0，而是1
这是这个字节是被编译器安插进去的一个字节，这样就使得这个空类的两个实例得以在内存中配置独一无二的地址。

2. 包含有virtual function的类的对象size至少是4字节(VPTR)

3. 对齐方式
复合数据类型里占用字节最多的元素与CPU对齐字节数中较小的一个，例子就是C1有int和char，size是8(两个int的大小)

4. 虚继承本质是有一个virtual base class table. 这个table一般是和virtual function table合并在一起的。
所以对于Class B6来说是以下的布局。

B6
----       A2 object's address [-1]
_vptr -->  type info           [0]
----       ~B6()'s address     [1]

所以B6和B4一样大。

5. 多重继承中,多个vtpr table, 子类的table与其中的一个virtual function table 合并,

所以D1的size是8,其中table _vtpr_C5是D1和C5共享的。
布局如下。

------         A2s object address [-1]
_vtpr_C5----  type info  [0]
               ~C's address [1]
------
               A2s object address [-1]
_vtpr_C6 ----  type info [0]
               ~C address [1]
------

Thanks for the detail description from Sunny.