关于c ++：dynamic_cast的好处/用途是什么？

What is the benefit/use of dynamic_cast?

本问题已经有最佳答案，请猛点这里访问。

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class Base
{
public:
virtual void print (){cout<<"Base"<<endl;}
};

class Derived :public Base
{
public:
virtual void print (){cout<<"Derived"<<endl;}
};

int main(){
Base *ptrb1=new Base();
//it will print Base
ptrb1->print();

Base *ptrb2=new Derived();
// it will print Derived
ptrb2->print();

Derived *ptrd=NULL;
//what is the difference between 2ways?
//what is the benefit of useing dynamic_cast?
ptrd=dynamic_cast<Derived*>(ptrb2);
ptrd->print();
}

如果我们可以通过添加虚函数使基类看到派生类的成员，并使基从派生类向obj指示，那么动态类型转换的好处(或两者之间的区别)是什么？

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如果不使用强制转换，则非虚拟成员函数将调用基方法。下面的代码显示了虚函数和非虚函数之间的区别，以及动态强制转换之后会发生什么。

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class Base
{
public:
void print2() {
cout <<"Base" << endl;
}
virtual void print() {
cout <<"Base" << endl;
}
};

class Derived : public Base
{
public:
void print2() {
cout <<"Derived" << endl;
}
virtual void print() {
cout <<"Derived" << endl;
}
};

int main()
{
Base* ptrb1 = new Base();
//it will print Base
ptrb1->print();
ptrb1->print2();

Base* ptrb2 = new Derived();
// it will print Derived
ptrb2->print();
ptrb2->print2();

Derived* ptrd = NULL;
//what is the difference between 2ways?
//what is the benefit of useing dynamic_cast?
ptrd = dynamic_cast<Derived*>( ptrb2 );
ptrd->print();
ptrd->print2();
}

如果您要使用static_cast(正如您在重新转换指针时可能倾向于认为可以这样做)，那么仍然从基派生的另一个类将被错误地转换。dynamic_cast将显示铸型不可能，而返回nullptr。