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)


: 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OutsideBase的隐式转换,而第二种选择涉及从OutsideDecorator的转换,这是“更好”的转换,因为OutsideBase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)
  { }

09-09 20:01
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