如何从中确定虚基类和派生类的大小？

Question

#include <iostream>
using namespace std;

class base1{};
class base2{virtual void show(){}};

class test1{    };
class test2{virtual void show(){}};

class derv1:public virtual base1{};
class derv12:public virtual base2{};

class derv2:public virtual base2, public test1{};
class derv22:public virtual base2, public virtual test1{};
class derv222:public virtual base2, public virtual test2{};

int main()
{
    cout<<"sizeof base1 = "<<sizeof(base1)<<endl;
    cout<<"sizeof base2 = "<<sizeof(base2)<<endl;
    cout<<"sizeof derv1 = "<<sizeof(derv1)<<endl;
    cout<<"sizeof derv12 = "<<sizeof(derv12)<<endl;

    cout<<"sizeof derv2 = "<<sizeof(derv2)<<endl;
    cout<<"sizeof derv22 = "<<sizeof(derv22)<<endl;
    cout<<"sizeof derv222 = "<<sizeof(derv222)<<endl;

}

输出:

sizeof base1 = 1
sizeof base2 = 4
sizeof derv1 = 4
sizeof derv12 = 8
sizeof derv2 = 8
sizeof derv22 = 8
sizeof derv222 = 12

我理解以下输出:

sizeof base1 : 1 => empty class , size is 1. result is OK.

sizeof base2 : 4 => since class has a virtual function, it adds 
                    a vitual pointer in each object of the class
                    ( 32 bit machine) , so added 4 bytes. Result is OK.

sizeof derv1 : 4 => base1 doesn't have  virtual function but since 
                    derv1 is virtually derived from base1 ,added 4 bytes. 
                    I think that each virtual base class
                    added a  pointer in object so i think Result is OK.     

sizeof derv12: 8 => 4 bytes( from virtual pointer) + 4 bytes
                     (virtual base class ) = 8 bytes. Result is OK.

在上述输出之后,我的困惑开始了

sizeof derv2  : 8 => 8 bytes( virtual pointer + virtually base class)
                    from base2 + 1 byte from test1 => 9 bytes and adds 3 padding 
                    bytes gives 12 bytes (should print). 
                     why "sizeof (test1)" is not adding in the output ??

sizeof derv22 : 8 => 8 bytes( virtual pointer + virtually base class)
                    from base2 + 4 byte (virtually base class) 
                    from test1  => 12 bytes (should print)

                    why 4 bytes from  test1 (virtual base class)  is not added 
                    in the output?? In the  size of(derv1 ) also has
                    a virtual base class( base1) and out put is 4 bytes means
                    they added 4 bytes in the output. 

sizeof derv222: 12 => 12 bytes( virtual pointer + virtually derived)
                      from base2 + 8 byte( virtual pointer + virtually derived)
                      from test2  => 16 bytes (should print)

我错过了一些或者这些尺寸是系统依赖还是其他任何东西？

Answer 1

sizeof(base1)和sizeof(test1)1 的原因仅仅是为了防止大多数派生对象的大小为0.这就是所有标准的禁止.允许基类子对象具有大小0(即,允许不占用任何字节),因此base1作为基础添加不一定必须添加任何类的大小.

编译器所做的优化,而不是为类型为空类的基类子对象分配任何字节,称为"空基类优化".标准不要求实现应用它,但实现不适合认真工作.

我认为derv22有点类似 - 如果编译器能够使用一个额外的指针处理两个虚拟基类,那么它有权这样做.因此,您可能只需要"支付"一次,而不是每个虚拟基数"支付".这可能取决于编译器以及类之间的确切关系,但是,我从未调查过不同的实现,以查看它们是否以及何时被迫添加多个指针值得开销.

显然derv222,至少对你的编译器来说已经做到了.我想这是因为base2和test2基类子对象需要单独的vtable指针.也许并不奇怪,如果你认为你的时候会发生什么static_cast一个derv222*为指针,以一个基地或其他-两个结果需要能够具有show()号召他们,并调用不同的功能(尽管该show功能目前什么都不做).我不知道这是否有可能为另一个编译器在8个字节实现这个继承-一个件事继承不具有使用虚函数表来实现的.

Answer 2

如何从中确定虚基类和派生类的大小？

使用sizeof().