【C++ Primer | 19】运行类型识别

运行类型识别

一、使用RTTI

dynamic_cast运算符的调用形式如下所示：

dynamic_cast<type*>(e)   //e是指针
dynamic_cast<type&>(e)   //e是左值
dynamic_cast<type&&>(e)  //e是右值

e能成功转换为type*类型的情况有三种：

1）e的类型是目标type的公有派生类：派生类向基类转换一定会成功。

2）e的类型是目标type的基类，当e是指针指向派生类对象，或者基类引用引用派生类对象时，类型转换才会成功，当e指向基类对象，试图转换为派生类对象时，转换失败。

3）e的类型就是type的类型时，一定会转换成功。

1. 测试代码：

 class Rect : public Shape {
 public:
     void fun() {};
     void prin() { cout << "我是方形" << endl; }
 };

 class Circle : public Shape {
 public:
     void fun() {};
     void prin() { cout << "我是圆形" << endl; }
 };

 void fun(Shape *p)
 {
     if (typeid(*p) == typeid(Rect))
     {
         Rect *r = dynamic_cast<Rect *>(p);
         r->prin();
     }
     if (typeid(*p) == typeid(Circle))
     {
         Circle *c = dynamic_cast<Circle *>(p);
         c->prin();
     }
 }
 int main()
 {
     Rect r;
     cout << "第一次执行fun函数：";
     fun(&r);

     Circle c;
     cout << "第二次执行fun函数：";
     fun(&c);

     return ;
 }

输出结果：

二、type_info类

1. 测试代码：

 #include <iostream>
 #include <typeinfo.h>
 using namespace std;

 class Base {
 public:
     virtual void vvfunc() {}
 };

 class Derived : public Base {};

 int main()
 {
     Derived* pd = new Derived;
     Base* pb = pd;
     cout << typeid(pb).name() << endl;   //prints "class Base *"
     cout << typeid(*pb).name() << endl;   //prints "class Derived"
     cout << typeid(pd).name() << endl;   //prints "class Derived *"
     cout << typeid(*pd).name() << endl;   //prints "class Derived"
     delete pd;
 }

输出结果：

2. 代码示例

 #include <iostream>
 #include <typeinfo>

 int main()
 {
     if (typeid(int).before(typeid(char)))
         std::cout << "int goes before char in this implementation.\n";
     else
         std::cout << "char goes before int in this implementation.\n";
 }

运行结果：

3. 代码示例

 #include <iostream>
 #include <typeinfo>
 #include <string>
 #include <utility>

 class person
 {
 public:

     person(std::string&& n) : _name(n) {}
     virtual const std::string& name() const { return _name; }

 private:

     std::string _name;
 };

 class employee : public person
 {
 public:

     employee(std::string&& n, std::string&& p) :
         person(std::move(n)), _profession(std::move(p)) {}

     const std::string& profession() const { return _profession; }

 private:
     std::string _profession;
 };

 void somefunc(const person& p)
 {
     if (typeid(employee) == typeid(p))
     {
         std::cout << typeid(p).name() << std::endl;
         std::cout << p.name() << " is an employee ";
         auto& emp = dynamic_cast<const employee&>(p);
         std::cout << "who works in " << emp.profession() << '\n';
     }
 }

 int main()
 {
     employee paul("Paul", "Economics");
     somefunc(paul);
 }

运行结果：

参考资料

• 记之c++相关：：dynamic_cast介绍
• std::type_info