c++ - 警告 : offset of on non-standard-layout type 'DerivedClass'

Question

我需要摆脱这个警告。据我了解，这似乎是因为 DerivedClass 不被视为普通旧数据。我读了关于 POD 的 cppreference和 Standard Layout Type但我仍然不明白为什么 DerivedClass 不是 POD。如果我将 BaseClass 的成员放入 DerivedClass 并且不使用继承 - 那么一切都很好。我使用 C++11

这是一个例子:

#include <iostream>

using namespace std;

class BaseClass
{
public:
    int a;
};

class DerivedClass : public BaseClass
{
public:
    int b;
};

int main(int argc, char *argv[])
{
    // warning: offset of on non-standard-layout type 'DerivedClass'
    int offset = offsetof(DerivedClass, b);

    cout << offset;
}

感谢任何帮助。

Answer 1

这些是标准布局类型的要求:

All non-static data members have the same access control

Has no virtual functions or virtual base classes

Has no non-static data members of reference type

All non-static data members and base classes are themselves standard layout types

直到 C++14:

Either

has no base classes with non-static data members, or

has no non-static data members in the most derived class and at most one base class with non-static data members

Has no base classes of the same type as the first non-static data member (see empty base optimization)

这部分适用于这个例子:

has no base classes with non-static data members,

在您的情况下，您有一个包含非静态数据成员的基类。删除 a 会删除警告。

C++14之后，也有类似的约束，两个类中不能有成员，只能有一个:

Has all non-static data members and bit-fields declared in the same class (either all in the derived or all in some base)