Question

由于空类的大小是1个字节但是这个空类实际上是继承的（在gcc编译器上）sizeof(derived1)的答案是8字节......，为什么会发生这种情况因为sizeof(derived1)必须是sizeof(char)+sizeof(virtual pointer)，这是1 + 4..ie 5字节。所以，从这个额外的3个字节到来的地方...... ???

#include <iostream>
using namespace std;

class Empty
{};

class Derived1 : virtual public Empty<br>
{
    char c;
};

int main()
{
    cout << "sizeof(Empty) " << sizeof(Empty) << endl;
    cout << "sizeof(Derived1) " << sizeof(Derived1) << endl;
    return 0;
}

Answer 1

就C ++标准而言，它都未指定，但通常虚拟继承会添加一个或两个指针反过来又施加了额外的对齐限制;的大小具有虚拟继承或虚函数的类几乎肯定必须是指针大小的倍数（大多数系统都是4或8）。

Answer 2

问题在于结构内存对齐。有关详细信息，请查看此atricle：http://en.wikipedia.org/wiki/Data_structure_alignment 正如您所看到的，内存填充取决于您的平台。

说明：

让我们尝试调试用于您班级的内存。我们知道变量不是虚拟。这意味着没有vintrual表的记录来确定使用哪个变量。因此，除非您没有虚函数，否则编译器不需要创建虚函数表。在我的机器上，虚函数表指针成员称为__vfptr。在调试时，除非添加虚函数，否则无法找到具有此名称的成员。这意味着您应该从计算中删除虚函数指针大小。尝试运行以下代码并阅读注释以获取详细信息：

struct Empty
{

};

struct VirtualNotEmpty: public virtual Empty
{
    char c;
};

struct VirtualNotEmptyEx: public virtual Empty
{
    char c1;
    char c2;
    char c3;
    char c4;
};

struct TrueVirtualNotEmpty: public virtual Empty
{
    virtual void f() {}
    char c;
};

void inspect_vne()
{
    VirtualNotEmpty vne;

    char* start_of_struct = (char*)(&vne);
    char* end_of_struct = (char*)(&vne);
    end_of_struct += sizeof(vne);
    char* start_of_var = (char*)(&(vne.c));

    printf("sizeof(VirtualNotEmpty): %d\n", sizeof(vne));                    // 8 bytes
    printf("Address of struct %p\n", start_of_struct);
    printf("Lenght of struct %d\n", (int)(end_of_struct - start_of_struct));    // same as sizeof
    printf("Address of var c %p\n", start_of_var);
    printf("Diff from start: %d\n", (int)(start_of_var - start_of_struct));        // 4 bytes: 1 byte for struct and 3 bytes for padding
    printf("Diff from end: %d\n", (int)(end_of_struct - start_of_var));            // 4 bytes:    1 byte for struct and 3 bytes for padding

    // [st][pd][pd][pd] [ch][pd][pd][pd]
    // st -- struct vt -- virtual table, pd -- padding byte ch -- char member variable
}

void inspect_tvne()
{
    TrueVirtualNotEmpty vne;

    char* start_of_struct = (char*)(&vne);
    char* end_of_struct = (char*)(&vne);
    end_of_struct += sizeof(vne);
    char* start_of_var = (char*)(&(vne.c));

    printf("sizeof(TrueVirtualNotEmpty): %d\n", sizeof(vne));                //    12
    printf("Address of struct %p\n", start_of_struct);
    printf("Lenght of struct %d\n", (int)(end_of_struct - start_of_struct)); // same as sizeof
    printf("Address of var c %p\n", start_of_var);
    printf("Diff from start: %d\n", (int)(start_of_var - start_of_struct)); // 8 bytes: 1 byte for struct + 3 bytes of struct padding + 4 bytes o vptr
    printf("Diff from end: %d\n", (int)(end_of_struct - start_of_var));        // 4 bytes: 1 byte for member variable + 3 bytes for padding

    // [st][pd][pd][pd] [vt][vt][vt][vt] [ch][pd][pd][pd]
    // st -- struct vt -- virtual table, pd -- padding byte ch -- char member variable
    // member variable TrueVirtualNotEmpty::c can't be packed at first byte, because of order of dectaled members __vptr is declared bevore
    // TrueVirtualNotEmpty::c
}

int main(int, char**)
{
    printf("Size of pointer %d", sizeof(int*)); // 4 on my machine    
    inspect_vne();
    inspect_tvne();
    printf("Size of VirtualNotEmptyEx %d\n", sizeof(VirtualNotEmptyEx));    // there is no need to create padding at
                                                                            // end for c1-c4 chars because there is space to pack them
    return 0;
}

正如您所看到的，我的机器上的结构大小应该可以被4分割。

为什么具有虚拟继承的派生类的大小会出乎意料？

2 个答案: