为什么我需要#include< typeinfo>使用typeid运算符时? [英] Why do I need to #include <typeinfo> when using the typeid operator?

问题描述

typeid 表示C ++ RTTI运算符也是C ++关键字.它返回一个 std::type_info 对象，其中包含(动态)类型特定的信息.

The typeid represents a C++ RTTI operator being also a C++ keyword. It returns a std::type_info object that holds (dynamic) type specific information.

据我从各种来源了解到，使用typeid时必须包含<typeinfo>，否则程序格式错误.实际上，如果我不包含上述标头，我的gcc5.2编译器甚至不会编译该程序.我不明白为什么要强制使用C ++ 关键字包含标头.我理解为每当我们使用在该标头中声明/定义的某个对象时都要强制标头，但typeid并非类类型.那么强制执行包含标头<typeinfo>的原因是什么?

From what I understood from various sources, one MUST include <typeinfo> when using typeid, otherwise the program is ill-formed. In fact, my gcc5.2 compiler doesn't even compile the program if I don't include the before-mentioned header. I don't understand why is a header inclusion mandated for the usage of a C++ keyword. I understand mandating a header for whenever we use some object declared/defined in that header, but typeid is not of a class type. So what is the reason behind this enforcement of including the header <typeinfo>?

推荐答案

下一段:

typeid表达式是指向对象的左值表达式 具有多态类型const的静态存储持续时间 std :: type_info或从其派生的某种类型.

The typeid expression is lvalue expression which refers to an object with static storage duration, of the polymorphic type const std::type_info or of some type derived from it.

因为它是一个左值表达式，它使用引用初始化来声明std::type_info的初始化程序. <typeinfo>包含该对象的定义.

Because it is an lvalue expression, which uses reference initialization to declare an initializer of std::type_info. <typeinfo> contains the definition for that object.

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