2010-06-16 173 views
1

我想在地图中保存提升信号对象(关联:信号名称→信号对象)。信号签名是不同的,所以第二种类型的地图应该是boost :: any将对象存储在数组中

map<string, any> mSignalAssociation; 

问题是如何在不定义新信号签名类型的情况下存储对象?

typedef boost::signals2::signal<void (int KeyCode)> sigKeyPressed; 
mSignalAssociation.insert(make_pair("KeyPressed", sigKeyPressed())); 
// This is what I need: passing object without type definition 
mSignalAssociation["KeyPressed"] = (typename boost::signals2::signal<void (int KeyCode)>()); 
// One more trying which won't work. And I don't want use this 
sigKeyPressed mKeyPressed; 
mSignalAssociation["KeyPressed"] = mKeyPressed; 

这一切tryings抛出错误:

/usr/include/boost/noncopyable.hpp: In copy constructor ‘boost::signals2::signal_base::signal_base(const boost::signals2::signal_base&)’: 
In file included from /usr/include/boost/signals2/detail/signals_common.hpp:17:0, 
/usr/include/boost/noncopyable.hpp:27:7: error: ‘boost::noncopyable_::noncopyable::noncopyable(const boost::noncopyable_::noncopyable&)’ is private 
/usr/include/boost/signals2/signal_base.hpp:22:5: error: within this context 
---------- 
/usr/include/boost/signals2/detail/signal_template.hpp: In copy constructor ‘boost::signals2::signal1<void, int&, boost::signals2::optional_last_value<void>, int, std::less<int>, boost::function<void(int)>, boost::function<void(const boost::signals2::connection&, int)>, boost::signals2::mutex>::signal1(const boost::signals2::signal1<void, int, boost::signals2::optional_last_value<void>, int, std::less<int>, boost::function<void(int)>, boost::function<void(const boost::signals2::connection&, int)>, boost::signals2::mutex>&)’: 
In file included from /usr/include/boost/preprocessor/iteration/detail/iter/forward1.hpp:52:0, 
/usr/include/boost/signals2/detail/signal_template.hpp:578:5: note: synthesized method ‘boost::signals2::signal_base::signal_base(const boost::signals2::signal_base&)’ first required here 
       from /usr/include/boost/signals2.hpp:16, 
--------- 
/usr/include/boost/signals2/preprocessed_signal.hpp: In copy constructor ‘boost::signals2::signal<void(int)>::signal(const boost::signals2::signal<void(int)>&)’: 
In file included from /usr/include/boost/signals2/signal.hpp:36:0, 
/usr/include/boost/signals2/preprocessed_signal.hpp:42:5: note: synthesized method ‘boost::signals2::signal1<void, int, boost::signals2::optional_last_value<void>, int, std::less<int>, boost::function<void(int)>, boost::function<void(const boost::signals2::connection&, int)>, boost::signals2::mutex>::signal1(const boost::signals2::signal1<void, int, boost::signals2::optional_last_value<void>, int, std::less<int>, boost::function<void(int)>, boost::function<void(const boost::signals2::connection&, int)>, boost::signals2::mutex>&)’ first required here 
       from /home/ockonal/Workspace/Projects/Pseudoform-2/include/Core/Systems.hpp:6, 
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我很困惑你的意思是:“问题是如何存储对象而不需要定义新的信号签名类型?” – GManNickG 2010-06-16 20:52:26

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@gman,我不想定义新类型的信号签名。在插入时直接使用它。 – Ockonal 2010-06-16 20:53:26

回答

4

这有什么好做anymap。提升信号简直是不可复制的。如果你想要一些可复制和自行清理的东西,你可以将它们包装在一个智能指针中,如shared_ptr

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如何在没有类型定义的情况下进行包装?我尝试过: 'shared_ptr keyEvent;' 但它说模板中错误的第一个参数。 – Ockonal 2010-06-17 07:33:30

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你不需要'typename'。它只需要依赖于模板类型参数的类型,但你的情况不是这样,并且'typename'在这里是禁止的。 – 2010-06-17 18:57:21