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。

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。

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)."

POD的概念和类型特征std::is_pod将在c++ 20中被弃用。有关进一步信息，请参阅此问题。

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