2011-05-11 73 views
0

时在文件A.hpp,我有“不匹配调用” 编译器错误使用boost ::信号

extern boost::signal<void (model::Bullet&, Point&, Point&, int)> signal_createBullet; 

等文件A.cpp,我有

boost::signal<void (model::Bullet&, Point&, Point&, int)> signal_createBullet; 

在文件B.hpp,我有一个类Entities具有静态成员函数receiveSignalCreateBullet,我想与signal_createBullet连接,像这样:(命名空间不再赘述)

class Entities 
{ 
    Entities() 
    { 
     signal_createBullet.connect(&receiveSignalCreateBullet); 
    } 

    public: 
     static void receiveSignalCreateBullet(const Bullet&, const Point&, const Point&, const int); 
}; 

inline static void receiveSignalCreateBullet(...) { ... } 

终于在文件C.cpp,我使用signal_createBullet这样的:

signal_createBullet(bullet, pos, bulletVector, count); 

甲乙编译成功(使用克++),但是C失败,此错误消息:

In member function ‘virtual void thrl::model::SingleStream::shoot(const thrl::utl::Point&, const thrl::utl::Point&, const thrl::utl::Point&) const’: 
src/Shot.cpp:25: error: no match for call to ‘(boost::signal4<void, thrl::model::Bullet&, thrl::utl::Point&, thrl::utl::Point&, int, boost::last_value<void>, int, std::less<int>, boost::function4<void, thrl::model::Bullet&, thrl::utl::Point&, thrl::utl::Point&, int> >) (const thrl::model::Bullet&, const thrl::utl::Point&, thrl::utl::Point&, int&)’ 
/usr/local/include/boost/signals/signal_template.hpp:330: note: candidates are: typename boost::signal4<R, T1, T2, T3, T4, Combiner, Group, GroupCompare, SlotFunction>::result_type boost::signal4<R, T1, T2, T3, T4, Combiner, Group, GroupCompare, SlotFunction>::operator()(T1, T2, T3, T4) [with R = void, T1 = thrl::model::Bullet&, T2 = thrl::utl::Point&, T3 = thrl::utl::Point&, T4 = int, Combiner = boost::last_value<void>, Group = int, GroupCompare = std::less<int>, SlotFunction = boost::function4<void, thrl::model::Bullet&, thrl::utl::Point&, thrl::utl::Point&, int>] 
/usr/local/include/boost/signals/signal_template.hpp:370: note:     typename boost::signal4<R, T1, T2, T3, T4, Combiner, Group, GroupCompare, SlotFunction>::result_type boost::signal4<R, T1, T2, T3, T4, Combiner, Group, GroupCompare, SlotFunction>::operator()(T1, T2, T3, T4) const [with R = void, T1 = thrl::model::Bullet&, T2 = thrl::utl::Point&, T3 = thrl::utl::Point&, T4 = int, Combiner = boost::last_value<void>, Group = int, GroupCompare = std::less<int>, SlotFunction = boost::function4<void, thrl::model::Bullet&, thrl::utl::Point&, thrl::utl::Point&, int>] 

虽然试图弄清楚这一点,我格式化我的电话,并在错误消息的第一候选人更容易对它们进行比较:

// my call 
‘(
    boost::signal 
    < 
     void 
     (
      thrl::model::Bullet&, 
      thrl::utl::Point&, 
      thrl::utl::Point&, 
      int 
     ), 
     boost::last_value<void>, 
     int, 
     std::less<int>, 
     boost::function 
     < 
      void 
      (
       thrl::model::Bullet&, 
       thrl::utl::Point&, 
       thrl::utl::Point&, 
       int 
      ) 
     > 
    > 
) 
(
    const thrl::model::Bullet&, 
    const thrl::utl::Point&, 
    thrl::utl::Point&, 
    int& 
)’ 

// what g++ expects 
typename boost::signal4<R, T1, T2, T3, T4, Combiner, Group, GroupCompare, SlotFunction>::result_type 
    boost::signal4<R, T1, T2, T3, T4, Combiner, Group, GroupCompare, SlotFunction>::operator()(T1, T2, T3, T4) 
    [ with 
     R = void, 
     T1 = thrl::model::Bullet&, 
     T2 = thrl::utl::Point&, 
     T3 = thrl::utl::Point&, 
     T4 = int, 
     Combiner = boost::last_value<void>, 
     Group = int, 
     GroupCompare = std::less<int>, 
     SlotFunction = boost::function 
     < 
      void 
      (
       thrl::model::Bullet&, 
       thrl::utl::Point&, 
       thrl::utl::Point&, 
       int 
      ) 
     > 
    ] 
// the second candidate is the same as the first, except that it's const 

除了候选人使用'便携'语法的事实(不,切换我的代码使用便携式风格没有区别)我看到两个电话之间没有区别,除了我呼叫的最后一个东西是int&候选人有一个int。我尝试从信号中去除int参数,看看这是否是问题,而事实并非如此。

任何人都明白为什么我会收到此错误?

回答

1

静态无效receiveSignalCreateBullet(常量子弹&,常量点&,常量点&,const int的);

为什么这里的参数是const?在信号声明中,它们不是常量。

+0

啊!他们最初是const的,虽然我在夸耀我的代码试图让事情发挥作用,但是我让它们不是const,并且忘记将它们改回来。感谢您为我侦察明显。 :) – Max 2011-05-11 22:30:01