---- 整理自狄泰软件唐佐林老师课程
在面向对象中可能出现下面的情况:
这个时候就会出现问题:(由于 赋值兼容性原则 )没法通过一个父类指针判断指向的是父类对象还是子类对象
基类指针是否可以强制类型转换为子类指针取决于动态类型
C++中如何得到 动态类型 ?
#include
#include using namespace std;class Base
{
public:virtual string type(){return "Base";}
};class Derived : public Base
{
public:string type(){return "Derived";}void printf(){cout << "I'm a Derived." << endl;}
};class Child : public Base
{
public:string type(){return "Child";}
};void test(Base* b)
{/* 危险的转换方式 */// Derived* d = static_cast(b);if( b->type() == "Derived" ){Derived* d = static_cast(b);d->printf();}// cout << dynamic_cast(b) << endl;
}int main(int argc, char *argv[])
{Base b;Derived d;Child c;test(&b);test(&d);test(&c);return 0;
}
向上/向下转换可参看 55 - 经典问题解析四(动态内存分配&虚函数&继承中的强制类型转换)
int i = 0;
const type_info& tiv = typeid(i);
const type_info& tii = typeid(int);
cout << (tiv == tii) << endl;
#include
#include
#include using namespace std;class Base
{
public:virtual ~Base(){}
};class Derived : public Base
{
public:void printf(){cout << "I'm a Derived." << endl;}
};void test(Base* b)
{const type_info& tb = typeid(*b);cout << tb.name() << endl;
}int main(int argc, char *argv[])
{int i = 0;const type_info& tiv = typeid(i);const type_info& tii = typeid(int);cout << (tiv == tii) << endl;Base b;Derived d;test(&b);test(&d);return 0;
}