c++ - 具有虚拟继承的类的大小

Question

我正在尝试使用https://www.onlinegdb.com/online_c++_compiler在下面的程序中进行虚拟继承。

#include <iostream>

using namespace std;

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; 
}; 

int main()
{
    cout << "Int is :" << sizeof(int) <<endl;
    cout << "Abase is :" << sizeof(ABase) << endl;
    cout << "Abase is :" << sizeof(BBase) << endl;
    cout << "Abase is :" << sizeof(CBase) << endl;
    cout << "Abase is :" << sizeof(ABCDerived) << endl;
    return 0;
}

ABase，BBase，CBase和ABCDerived的大小分别为4、16、16和40。
通常，根据我的阅读，我期望BBase和CBase为12字节(两个int变量为8字节，vPtr为4字节)。但是我得到16，即使我假设vPtr使用的是8字节，ABCDerived的大小也为40字节？
OnlineGBD工具声称使用G++ / C++编译器。
我知道以前在此论坛中曾问过类似的问题，但我无法理解此大小的计算。

Answer 1

如果您使用#pragma pack(1)删除填充，以减少混淆的空间，请考虑:

#include <iostream>

using namespace std;
#pragma pack(1)

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; 
}; 

int main()
{
    cout << "void* is :" << sizeof(void*) << endl;
    cout << "Int is :" << sizeof(int) <<endl;
    cout << "Abase is :" << sizeof(ABase) << endl;
    cout << "Bbase is :" << sizeof(BBase) << endl;
    cout << "Cbase is :" << sizeof(CBase) << endl;
    cout << "ABCDerived is :" << sizeof(ABCDerived) << endl;
}

这将为您提供输出:

void* is :8
Int is :4
Abase is :4
Bbase is :16
Cbase is :16
ABCDerived is :32

Bbase和 Cbase包含:

一个Abase虚拟基础(4个字节)

指向其Abase虚拟基础的指针(8字节)

自己的int成员(4个字节)

总计= 16个字节
ABCDerived包含:

Bbase基础(16字节)

一个Cbase基础(只需要存储int成员+一个指向通过Abase = 4 + 8 = 12字节继承的现有Bbase虚拟基础的指针)

自己的int成员(4个字节)

总计= 32字节