【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 0;
}

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.
";
    else
        std::cout << "char goes before int in this implementation.
";
}

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() << '
';
    }
}

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