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我在尝试使用装饰器模式时遇到问题。构造函数正在打印出用于调试的地址。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;
}

我的问题出在 Group::Group() 中。我相信用未初始化的 _outside 初始化 Group 的 Decorator 基础部分很好 - Decorator 唯一想要的是指向对象的指针。我的问题是 Decorator(_outside) 似乎正在调用我不想要的复制构造函数。

gdb 善良：

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>}}

我在 g1 的构造函数之前中断并将几个 _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 将 _outside._decorated 的未初始化值作为其 _decorated 成员，这意味着复制构造函数已被调用。如果我将复制构造函数代码添加到类装饰器：

Decorator(const Decorator & that)
{   std::cout << "Copy constructor - this: " << this << " - that: " << &that << std::endl;   }

它实际上调用它。

如果我将 Group 构造函数的第二行从

: Decorator(_outside)

到

: 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>}}

装饰器复制构造函数没有被调用，一切似乎都很好。我不喜欢这个解决方案，因为它要求下游的每个班级都记得这样做。

有什么方法可以从 Decorator 派生 Group 并使用 Decorator 成员而不调用复制构造函数？

4

1 回答 1

1

我的问题是 Decorator(_outside) 似乎正在调用我不想要的复制构造函数。

你期望它做什么？

Decorator没有构造函数，Outside因此符合条件的构造函数是：

Decorator(const Base&)

或隐式定义的复制构造函数：

Decorator(const Decorator&)

第一个选项涉及从Outsideto的隐式转换，Base而第二个选项涉及从 to 的转换Outside，Decorator这是一个“更好”的转换，因为OutsidetoBase转换“通过”Decorator到 get Base。

正如您所发现的，要调用您想要的构造函数，您需要显式地进行所需的转换：

Decorator(static_cast<Base&>(_outside))

这是必要的，因为您传递的类型确实是aDecorator所以它当然更喜欢复制构造函数。

另一种解决方案是添加一个构造函数来代替复制构造函数，例如一个模板，适当地约束：

template<typename T>
  Decorator(const T& decorated, typename boost::enable_if<boost::is_base_of<T, Base> >* = 0)
  : _decorated(&decorated)
  { }

这将用于派生于Base但不是 aBase且不是 a 的任何内容Decorator

在 C++11 中，你可以让它更干净一些

template<typename T,
         typename Requires = typename std::enable_if<std::is_base_of<T, Base>::value>>
  Decorator(const T& decorated)
  : _decorated(&decorated)
  { }
于 2013-04-01T21:13:37.083 回答