在运行时从没有虚拟函数的基类方法访问派生类方法
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我有以下代码示例。一个基类和两个派生类具有各自的功能（分别为function1和function2）。在基类中，function1和function2都不是虚拟的。我无法更改此设置，因为这些类已经实现。
#include <iostream>
class Base
{
public:
Base(){}
Base(int val) : m_base(val){}
virtual ~Base(){}
//base class methods
private:
int m_base;
};
class Derived1 : public Base
{
public:
Derived1(int val) : m_derived1(val){}
~Derived1(){}
void print1(){std::cout << \"Using Derived 1 class method\" << std::endl;};
private:
int m_derived1;
};
class Derived2 : public Base
{
public:
Derived2(int val) : m_derived2(val){}
~Derived2(){}
void print2(){std::cout << \"Using Derived 2 class method\" << std::endl;};
private:
int m_derived2;
};
int main()
{
int option;
std::cin >> option;
Base* b = new Base(5);
Derived1* d1 = new Derived1(5);
Derived2* d2 = new Derived2(5);
d1->print1(); //obviously this works
d2->print2(); //obviously this works
//In reality I thus have a function in d1 and d2 which is not present in b
//I have to decide on runtime which class I need to use
if(option == 1)
{
b = d1;
}
else if(option == 2)
{
b = d2;
}
/*
Rest of program ...
b->invokeMethod;
// ... //
b->invokeMoreMethods;
*/
//Call derived functions on base object
//b->print1(); //fails obviously
if(option == 1)
{
dynamic_cast<Derived1*>(b)->print1(); //will work if option == 1 is specified (*)
}
if(option == 2)
{
dynamic_cast<Derived2*>(b)->print2(); //will work if option == 2 is specified (*)
}
return 0;
}
没有找到相关结果
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1 个回复
凰葱崎济邯
int main() { Base * b = 0; // You were leaking a Base object here boost::function< void () > f; ... if ( option == 1 ) { Derived1 d1 = new Derived1; // Only instantiate the object you will use b = d1; f = boost::bind( &Derived1::print1, d1 ); } else if ( option == 2 ) { Derived2 d2 = new Derived2; b = d2; f = boost::bind( &Derived2::print2, d2 ); } ... f(); // will call either d1->print1(), or d2->print2() ... delete b; // should really use a smart pointer but I left // the raw pointer to minimize changes }