我遇到过几次POD-type这个术语。 这是什么意思?
当前回答
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。
其他回答
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
POD(普通旧数据)对象具有这些数据类型中的一种——基本类型、指针、联合、结构、数组或类——没有构造函数。相反,非pod对象是存在构造函数的对象。POD对象在获得与其类型对应的适当大小的存储空间时开始其生命周期,在对象的存储空间被重用或释放时结束其生命周期。
PlainOldData类型也不能包含:
虚函数(它们自己的或继承的) 虚拟基类(直接或间接)。
PlainOldData更宽松的定义包括带有构造函数的对象;但不包括那些虚拟的东西。PlainOldData类型的重要问题是它们是非多态的。继承可以用POD类型完成,但是它应该只用于ImplementationInheritance(代码重用),而不是多态性/子类型。
一个常见的(虽然不是严格正确的)定义是PlainOldData类型是没有VeeTable的任何类型。
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)."
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.
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。
推荐文章
- 为什么这个结合赋值和相等检查的if语句返回true?
- cplusplus.com给出的错误、误解或坏建议是什么?
- 找出质数最快的算法是什么?
- time_t最终的类型定义是什么?
- c++枚举类可以有方法吗?
- 使arrayList.toArray()返回更具体的类型
- 格式化IO函数(*printf / *scanf)中的转换说明符%i和%d之间的区别是什么?
- 将析构函数设为私有有什么用?
- main()中的Return语句vs exit()
- 为什么c#不提供c++风格的'friend'关键字?
- 在函数的签名中添加关键字
- 我如何在Visual Studio中预处理后看到C/ c++源文件?
- 为什么在标准容器中使用std::auto_ptr<>是错误的?
- 用比较double和0
- 保护可执行文件不受逆向工程的影响?