我最近遇到以下问题。我有一个具有Base_encapsulated_class的基类。
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为我提供了
m_encapsulated_class
的两个实例,这在c ++中显然是合法的(我感到有些惊讶)。但这不是我想要的。我在派生类中需要一个
m_encapsulated_class
成员。那么可以在基类中扩展复合成员吗? 最佳答案
您可以使用虚拟getter强制成员继承:
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;
};