POD代表普通旧数据——也就是说，一个没有构造函数、析构函数和虚成员函数的类(无论是用关键字struct还是用关键字class定义的)。维基百科关于POD的文章更详细一些，并将其定义为:

c++中的普通旧数据结构是一个聚合类，它只包含PODS作为成员，没有用户定义的析构函数，没有用户定义的复制赋值操作符，也没有指针到成员类型的非静态成员。

更多的细节可以在c++ 98/03的答案中找到。c++ 11改变了围绕POD的规则，极大地放松了它们，因此需要在这里进行后续回答。

据我所知，POD (PlainOldData)只是一个原始数据-它不需要:

要被建造， 被毁灭， 拥有自定义操作符。 必须没有虚函数， 并且不能重写操作符。

如何检查某物是否是POD?有一个结构体叫做std::is_pod:

namespace std {
// Could use is_standard_layout && is_trivial instead of the builtin.
template<typename _Tp>
  struct is_pod
  : public integral_constant<bool, __is_pod(_Tp)>
  { };
}

(来自头文件type_traits)

参考:

http://en.cppreference.com/w/cpp/types/is_pod http://en.wikipedia.org/wiki/Plain_old_data_structure http://en.wikipedia.org/wiki/Plain_Old_C++_Object 文件type_traits

With C++, Plain Old Data doesn't just mean that things like int, char, etc are the only types used. Plain Old Data really means in practice that you can take a struct memcpy it from one location in memory to another and things will work exactly like you would expect (i.e. not blow up). This breaks if your class, or any class your class contains, has as a member that is a pointer or a reference or a class that has a virtual function. Essentially, if pointers have to be involved somewhere, its not Plain Old Data.

非常非正式地:

POD是一种类型(包括类)，c++编译器保证结构中不会发生“魔法”:例如，指向虚表的隐藏指针，转换为其他类型时应用于地址的偏移量(至少如果目标也是POD)，构造函数或析构函数。粗略地说，当类型中只有内置类型和它们的组合时，类型就是POD。结果是某种“像”C类型的东西。

非正式地:

int, char, wchar_t, bool, float, double are PODs, as are long/short and signed/unsigned versions of them. pointers (including pointer-to-function and pointer-to-member) are PODs, enums are PODs a const or volatile POD is a POD. a class, struct or union of PODs is a POD provided that all non-static data members are public, and it has no base class and no constructors, destructors, or virtual methods. Static members don't stop something being a POD under this rule. This rule has changed in C++11 and certain private members are allowed: Can a class with all private members be a POD class? Wikipedia is wrong to say that a POD cannot have members of type pointer-to-member. Or rather, it's correct for the C++98 wording, but TC1 made explicit that pointers-to-member are POD.

形式上(c++ 03标准):

3.9(10): "Arithmetic types (3.9.1), enumeration types, pointer types, and pointer to member types (3.9.2) and cv-qualified versions of these types (3.9.3) are collectively caller scalar types. Scalar types, POD-struct types, POD-union types (clause 9), arrays of such types and cv-qualified versions of these types (3.9.3) are collectively called POD types" 9(4): "A POD-struct is an aggregate class that has no non-static data members of type non-POD-struct, non-POD-union (or array of such types) or reference, and has no user-define copy operator and no user-defined destructor. Similarly a POD-union is an aggregate union that has no non-static data members of type non-POD-struct, non-POD-union (or array of such types) or reference, and has no user-define copy operator and no user-defined destructor. 8.5.1(1): "An aggregate is an array or class (clause 9) with no user-declared constructors (12.1), no private or protected non-static data members (clause 11), no base classes (clause 10) and no virtual functions (10.3)."

普通旧数据

简而言之，它是所有内置数据类型(例如int、char、float、long、unsigned char、double等)和POD数据的所有聚合。是的，这是递归定义。；）

更清楚地说，POD就是我们所说的“结构体”:仅存储数据的单元或单元组。

POD代表普通旧数据——也就是说，一个没有构造函数、析构函数和虚成员函数的类(无论是用关键字struct还是用关键字class定义的)。维基百科关于POD的文章更详细一些，并将其定义为:

c++中的普通旧数据结构是一个聚合类，它只包含PODS作为成员，没有用户定义的析构函数，没有用户定义的复制赋值操作符，也没有指针到成员类型的非静态成员。

更多的细节可以在c++ 98/03的答案中找到。c++ 11改变了围绕POD的规则，极大地放松了它们，因此需要在这里进行后续回答。