我最近遇到了以下问题。我有一个Base类,它有一个Base_encapsulated_class。
class Base_class
{
public:
Base_class(int int_value, short short_value) :
m_encapsulated_class(int_value, short_value)
{};
~Base_class(void){};
Base_encapsulated_class m_encapsulated_class;
};
class Base_encapsulated_class
{
public:
Base_encapsulated_class(int int_value, short short_value)
: m_int(int_value),
m_short(short_value)
{};
~Base_encapsulated_class(void){};
private:
int m_int;
short m_short;
};
现在我想扩展Base_class
和Base_encapsulated_class
来实现:
class Derived_class : public Base_class
{
public:
Derived_class(int int_value, short short_value, bool boolean_value) :
Base_class(int_value, short_value)
{};
~Derived_class(void)
{};
Derived_encapsulated_class m_encapsulated_class;
};
class Derived_encapsulated_class : public Base_encapsulated_class
{
public:
Derived_encapsulated_class(int int_value, short short_value, bool boolean_value) :
Base_encapsulated_class(int_value, short_value),
m_bool(boolean_value)
{
}
~Derived_encapsulated_class(void)
{
}
private:
bool m_bool;
};
这种方法在派生类对象中给出了两个m_encapsulated_class
实例,这显然在c ++中是合法的(我有点惊讶)。
但这不是我想要的。我想在我的派生类中有一个m_encapsulated_class
成员。那么可以在基类中扩展复合成员吗?
答案 0 :(得分:1)
您可以使用虚拟getter强制继承成员:
class Base_class
{
public:
Base_class(int int_value, short short_value) :
m_encapsulated_class(int_value, short_value)
{};
protected:
virtual Base_encapsulated_class Get_encapsulated_class() /* = 0 if you want to enforce encapsulated derivation */
{
return m_encapsulated_class;
}
private:
Base_encapsulated_class m_encapsulated_class;
};
class Derived_class : public Base_class
{
public:
Derived_class(int int_value, short short_value, bool boolean_value) :
Base_class(int_value, short_value)
{};
~Derived_class(void)
{};
protected:
virtual Base_encapsulated_class Get_encapsulated_class()
{
return m_derived_encapsulated_class;
}
private:
Derived_encapsulated_class m_derived_encapsulated_class;
};