我在尝试使用装饰器模式时遇到问题。构造函数正在打印地址以进行调试。编译:
g++ -g -o go Decorator.cpp
我的简化代码:
#include <iostream>
class Base
{
public:
Base()
{
std::cout << "Base created - this: " << this << std::endl;
}
virtual ~Base() {}
};
class Decorator : public Base
{
public:
Decorator(const Base & decorated)
: _decorated(&decorated)
{
std::cout << "Decorator created - this: " << this << " created - _decorated is " << _decorated << std::endl;
}
~Decorator()
{
std::cout << "Decorator destroyed" << std::endl;
std::cout << " This: " << this << ", _decorated: " << _decorated << std::endl;
}
private:
const Base * _decorated;
};
class Inside : public Base
{
public:
Inside()
{ std::cout << "Inside created - this: " << this << std::endl; }
};
class Outside : public Decorator
{
public:
Outside(const Base & decorated)
: Decorator(decorated)
{
std::cout << "Outside created - this: " << this << std::endl;
}
};
class Group : public Decorator
{
public:
Group()
: Decorator(_outside)
, _outside(_inside)
{
std::cout << "Group created - this: " << this << std::endl;
}
~Group()
{
std::cout << "Group destroyed" << std::endl;
}
private:
Inside _inside;
Outside _outside;
};
int main()
{
std::cout << "Hi there" << std::endl;
Group g1;
std::cout << "Done" << std::endl;
}
我的问题在Group::Group()中。我相信用未初始化的_outside初始化Group的Decorator基础部分是好的-Decorator唯一想要的就是指向该对象的指针。我的问题是Decorator(_outside)似乎正在调用复制构造函数,而我不希望这样做。
gdb的优点:
Breakpoint 1, _fu0___ZSt4cout () at Decorator.cpp:63
63 Group g1;
(gdb) print g1
$1 = {<Decorator> = {<Base> = {_vptr.Base = 0x77c34e29},
_decorated = 0x77c34e42}, _inside = warning: can't find linker symbol for vi
rtual table for `Inside' value
{<Base> = {
_vptr.Base = 0x401a90}, <No data fields>},
_outside = {<Decorator> = {<Base> = {_vptr.Base = 0x22ff58},
_decorated = 0x401af6}, <No data fields>}}
我先打破g1的构造函数,然后将几个_decorated成员写入已知值以帮助调试。
(gdb) set g1._decorated = 0
(gdb) set g1._outside._decorated = 0xeeeeeeee
(gdb) print g1
$2 = {<Decorator> = {<Base> = {_vptr.Base = 0x77c34e29}, _decorated = 0x0},
_inside = warning: can't find linker symbol for virtual table for `Inside' val
ue
{<Base> = {_vptr.Base = 0x401a90}, <No data fields>},
_outside = {<Decorator> = {<Base> = {_vptr.Base = 0x22ff58},
_decorated = 0xeeeeeeee}, <No data fields>}}
(gdb) n
Base created - this: 0x22ff34
Inside created - this: 0x22ff34
Base created - this: 0x22ff38
Decorator created - this: 0x22ff38 created - _decorated is 0x22ff34
Outside created - this: 0x22ff38
Group created - this: 0x22ff2c
65 std::cout << "Done" << std::endl;
(gdb) print g1
$3 = {<Decorator> = {<Base> = {_vptr.Base = 0x4042b8},
_decorated = 0xeeeeeeee}, _inside = {<Base> = {
_vptr.Base = 0x4042c8}, <No data fields>},
_outside = {<Decorator> = {<Base> = {_vptr.Base = 0x4042d8},
_decorated = 0x22ff34}, <No data fields>}}
在构造函数之后,g1._decorated具有未初始化的_outside._decorated值作为其_decorated成员,这意味着已调用复制构造函数。如果我将复制构造函数代码添加到类Decorator中:
Decorator(const Decorator & that)
{ std::cout << "Copy constructor - this: " << this << " - that: " << &that << std::endl; }
实际上确实如此。
如果我将Group构造函数的第二行从
: Decorator(_outside)
至
: Decorator(static_cast<const Base &>(_outside))
并运行gdb
Breakpoint 1, _fu0___ZSt4cout () at Decorator.cpp:63
63 Group g1;
(gdb) print g1
$1 = {<Decorator> = {<Base> = {_vptr.Base = 0x77c34e29},
_decorated = 0x77c34e42}, _inside = warning: can't find linker symbol for vi
rtual table for `Inside' value
{<Base> = {
_vptr.Base = 0x401a90}, <No data fields>},
_outside = {<Decorator> = {<Base> = {_vptr.Base = 0x22ff58},
_decorated = 0x401af6}, <No data fields>}}
(gdb) set g1._decorated = 0
(gdb) set g1._outside._decorated = 0xeeeeeeee
(gdb) print g1
$2 = {<Decorator> = {<Base> = {_vptr.Base = 0x77c34e29}, _decorated = 0x0},
_inside = warning: can't find linker symbol for virtual table for `Inside' val
ue
{<Base> = {_vptr.Base = 0x401a90}, <No data fields>},
_outside = {<Decorator> = {<Base> = {_vptr.Base = 0x22ff58},
_decorated = 0xeeeeeeee}, <No data fields>}}
(gdb) n
Base created - this: 0x22ff2c
Decorator created - this: 0x22ff2c created - _decorated is 0x22ff38
Base created - this: 0x22ff34
Inside created - this: 0x22ff34
Base created - this: 0x22ff38
Decorator created - this: 0x22ff38 created - _decorated is 0x22ff34
Outside created - this: 0x22ff38
Group created - this: 0x22ff2c
65 std::cout << "Done" << std::endl;
(gdb) print g1
$3 = {<Decorator> = {<Base> = {_vptr.Base = 0x4042b8},
_decorated = 0x22ff38}, _inside = {<Base> = {
_vptr.Base = 0x4042c8}, <No data fields>},
_outside = {<Decorator> = {<Base> = {_vptr.Base = 0x4042d8},
_decorated = 0x22ff34}, <No data fields>}}
装饰器副本构造函数未调用,并且一切似乎都很好。我不喜欢这种解决方案,因为它要求下游的每个类都记住要这样做。
有什么方法可以从带有成员Decorator的Decorator派生Group而不调用副本构造函数?
最佳答案
您期望它做什么?Decorator
没有采用Outside
的构造函数,因此符合条件的构造函数为:
Decorator(const Base&)
或隐式定义的副本构造函数:
Decorator(const Decorator&)
第一种选择涉及从
Outside
到Base
的隐式转换,而第二种选择涉及从Outside
到Decorator
的转换,这是“更好”的转换,因为Outside
到Base
的转换“经过”了Decorator
才能到达Base
。如您所见,要调用构造函数,您需要明确地进行所需的转换:
Decorator(static_cast<Base&>(_outside))
这是必需的,因为您要传递的类型实际上是
Decorator
,因此,它当然更喜欢复制构造函数。另一种解决方案是添加将要使用的构造函数代替复制构造函数,例如适当限制的模板:
template<typename T>
Decorator(const T& decorated, typename boost::enable_if<boost::is_base_of<T, Base> >* = 0)
: _decorated(&decorated)
{ }
这将用于从
Base
派生但不是Base
也不是Decorator
的任何内容在C++ 11中,您可以使其更加整洁
template<typename T,
typename Requires = typename std::enable_if<std::is_base_of<T, Base>::value>>
Decorator(const T& decorated)
: _decorated(&decorated)
{ }