c++ - STL std::map 和 std::vector ；检查 map 中的对象类型

Question

所以我在我的程序中进行事实检查时遇到了问题，代码如下: 该 vector 包含 3 种类型的派生对象，我只想要 vector 中每个基本对象的小计。我似乎也找不到合适的语法。

class Base{
virtual void method() = 0;
}  

class derived_1 : public Base{
    virtual void method();
}
class derived_2 : public Base{
    virtual void method();
}
class derived_3 : public Base{
    virtual void method();
}
class general_class{
private: 
    //objects of derived types have been instantiated into the vector already
    map<string,vector<Base*>> base_map;

    void print(){
        //This line prints the key and size
        cout << iter->first << "    " << iter->.size();

        int d1_count = 0, d2_count = 0,d3_count = 0;

        for(iter=accounts_map.begin();iter !=accounts_map.end();iter++){

            //So I know that the loop iterates through the map
            //how do I fact check to determine which object was found?
            //The below code is incorrect

            if(iter->second[i] == (derived_1 /*"objects"*/)){
                d1_count++;
            }
            if(iter->second[i] == (derived_2 /*"objects"*/)){
                d2_count++;
            }
            if(iter->second[i] == (derived_3 /*"objects"*/)){
                d3_count++;
            }
        }
    }

}

我不确定语法是什么或检查正确对象类型背后的逻辑。

Answer 1

有很多方法可以实现您的目标。您可以扩展您的 Base 接口(interface)以返回一些对象种类标识符。另一种选择是使用 RTTI:

for(auto pObj : vector)
{
  if(dynamic_cast<derived1*>(pObj))
    d1_count++;
}

另请注意，您的接口(interface)基类定义不正确。您必须提供虚拟析构函数，否则不会调用派生类的析构函数。正确的版本应该是:

class Base{
    virtual void method() = 0;
    virtual ~Base() {};
}