我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的基类:
class interface {
public:
virtual interface(){}
virtual ~interface(){}
};
从中派生出两个更抽象的类。
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(){}
};
以及来自二级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?像往常一样,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呢?
vector</* What should I put in here? */> interface_vector;
只push_back新的derived_1
实例和derived_2
到一个通用向量vector<interface*>
是没有问题的,因为它们有interface
类作为祖先。
顺便说一句:你不需要derived_1
类和derived_2
类再次从interface
继承。这是不常见的,我很确定这可能会导致其他问题。
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]));
只需添加上面的循环即可。重点是,您可以投射到interface*
这是您的vector<interface*>
所期望的。