将抽象派生类对象存储在基类向量中

Question

我已经学会了通过存储基类指针将派生类指针存储在基类向量中：

vector<base*> base_vector;
base_vector.push_back(new derived());
// free memory at the end

但是如果我有一个抽象基类：

class interface {
public:
    virtual interface(){} 
    virtual ~interface(){}
};

从中派生出另外两个抽象类。

class abstract_derived_1 : virtual public interface 
{ 
public:
    virtual abstract_derived_1(){} 
    virtual ~abstract_derived_1(){}
};

class abstract_derived_2 : virtual public interface 
{ 
public:
    virtual abstract_derived_2(){} 
    virtual ~abstract_derived_2(){}
};

以及来自次要抽象类的其他几个派生类：

class derived_1 : virtual public interface, virtual public abstract_derived_1
{
private:
    double value;
public:
    derived_1(){value=0;}
    derived_1(const double val1, const double val2) { value = val1+val2; }
    ~derived_1(){}
};


class derived_2 : virtual public interface, virtual public abstract_derived_2
{
private:
    string name;
public:
    derived_2(){name="";}
    derived_2(string my_str) { name = my_str; }
};

是否可以将它们全部存储在多态向量中？ 和往常一样，我做了以下工作：

vector<abstract_derived_1*> abs1;
vector<abstract_derived_2*> abs2;
abs1.push_back(new derived_1(1,2));
abs2.push_back(new derived_2("polymorphism"));

但是如何将两个多态向量存储在基类向量中？

vector</* What should I put in here? */> interface_vector;

Answer 1

只需将新derived_1和derived_2实例derived_2回通用向量vector<interface*>就没有问题，因为它们具有interface类作为祖先。

顺便说一句：您无需从interface再次继承derived_1类和derived_2类。 这是罕见的，我很确定这可能会导致其他问题。

Answer 2

vector<interface*> interface_vector;
// Loop through abs1 with an index of i
interface_vector.push_back(dynamic_cast<interface*>(abs1[i]));
// Loop through abs2 with an index of i
interface_vector.push_back(dynamic_cast<interface*>(abs2[i]));

只需在上面添加循环即可。 主要的一点是，你可以转换为interface*这就是你的vector<interface*>的期望。