我有两个课程B
和Y
我无法按要求更改或编辑。它们的功能相同,但名称不同。
我希望有一个通用接口,在运行时选择类,具体取决于下面代码中描述的某个输入变量。我不确定应该使用哪种设计模式。如何创建WrapperYB
类,根据创建的对象选择Y::show
或B::showing
。
class A
{
public:
A() {}
virtual ~A();
virtual void show() { cout << "show A" << endl;}
};
class B:A
{
public:
B() {}
virtual ~B();
virtual void show() { cout << "show B" << endl;}
};
class X
{
char m_i;
public:
Y() { m_i = 'X';}
virtual void showing() { cout << "showing " << m_i << endl;}
};
class Y:X
{
public:
Y() { m_i = 'Y';}
virtual void showing() { cout << "showing " << m_i << endl;}
};
class WrapperYB
{
// to be implemented
public:
explicit WrapperYB(const int& type);
void show();
};
int main(){
WrapperYB objY(1);
objY.show(); // must call Y::showing
WrapperYB objB(0);
objB.show(); // must call B::show
}
答案 0 :(得分:3)
如果您的编译器支持C ++ 17 Standard,您可以使用std::variant
尝试此解决方案。这与@Nicolas的答案中的解决方案类似,但是variant
将为您处理实现细节,不会使用动态内存分配,并且支持复制和赋值等其他内容。
#include <variant>
#include <utility>
#include <type_traits>
class WrapperYB {
public:
using variant_type = std::variant<Y, B>;
template <typename... Args,
std::enable_if_t<std::is_constructible_v<variant_type, Args...>>* = nullptr>
WrapperYB(Args&& ... args) : m_variant(std::forward<Args>(args)...) {}
variant_type& variant() noexcept { return m_variant; }
const variant_type& variant() const noexcept { return m_variant; }
void show()
{ std::visit(ShowImpl{}, m_variant); }
private:
struct ShowImpl {
void operator() (Y& y) const { y.showing(); }
void operator() (B& b) const { b.show(); }
};
variant_type m_variant;
};
See the full working example on coliru.
您可以通过让包含std::unique_ptr<A>
或std::unique_ptr<X>
包含来概括包装。
答案 1 :(得分:2)
我在提议:
#include <iostream>
using namespace std;
class A
{
public:
A() {}
virtual ~A() {}
virtual void show() { cout << "show A" << endl;}
};
class B:A
{
public:
B() {}
virtual ~B() {}
virtual void show() { cout << "show B" << endl;}
};
class X
{
protected:
char m_i;
public:
X () { m_i = 'X';}
virtual void showing() { cout << "showing " << m_i << endl;}
};
class Y:X
{
public:
Y() { m_i = 'Y';}
virtual void showing() { cout << "showing " << m_i << endl;}
};
class WrapperYB
{
public:
enum class Which { B, Y };
public:
explicit WrapperYB (int n)
: which(Which(n))
{
switch (which)
{
case Which::B: ptr.b = new B; break;
case Which::Y: ptr.y = new Y; break;
}
}
~WrapperYB ()
{
switch (which)
{
case Which::B: delete ptr.b; break;
case Which::Y: delete ptr.y; break;
}
}
WrapperYB (const WrapperYB&) = delete;
WrapperYB& operator = (const WrapperYB&) = delete;
public:
void show()
{
switch (which)
{
case Which::B: ptr.b->show() ; break;
case Which::Y: ptr.y->showing(); break;
}
}
private:
Which which;
union {
Y* y;
B* b;
} ptr;
};
int main(){
WrapperYB objY(1);
objY.show(); // must call Y::showing
WrapperYB objB(0);
objB.show(); // must call B::show
}
这不是“香草”的设计模式,我不认为,更多的是适配器和歧视联合的组合。
请注意,无法按原样复制或分配WrapperYB。
答案 2 :(得分:2)
您可以将标准虚拟分派方法与抽象基本适配器类和子类一起使用,以用于所需的每种对象类型。使用工厂方法创建对象。
#include <memory>
//pre-defined structures Y, B
struct Y
{
Y(){}
~Y(){}
void show(){}
};
struct B
{
B(){}
~B(){}
void showing(){}
};
// Abstract adaptor base class.
struct Adaptor
{
virtual void show() = 0;
};
// A subclass of Adaptor for each type of object to be wrapped.
struct Adaptor_Y: Adaptor
{
Adaptor_Y(): y(){}
void show() override
{
y.show();
}
private:
Y y;
};
struct Adaptor_B: Adaptor
{
Adaptor_B(): b(){}
void show() override
{
b.showing();
}
private:
B b;
};
// Factory method constructs the proper object and returns a pointer.
std::unique_ptr<Adaptor> get_adaptor(int flag)
{
if(flag == 0)
{
return std::make_unique<Adaptor_B>();
}
else if(flag == 1)
{
return std::make_unique<Adaptor_Y>();
}
else throw std::runtime_error("Invalid flag value");
}