在文件A.hpp中,我有
extern boost::signal<void (model::Bullet&, Point&, Point&, int)> signal_createBullet;
所以在文件A.cpp中,我有
boost::signal<void (model::Bullet&, Point&, Point&, int)> signal_createBullet;
在文件B.hpp中,我有一个类Entities
,它有一个静态成员函数receiveSignalCreateBullet
我希望与signal_createBullet
连接,如下所示:(为简洁省略了名称空间)
class Entities
{
Entities()
{
signal_createBullet.connect(&receiveSignalCreateBullet);
}
public:
static void receiveSignalCreateBullet(const Bullet&, const Point&, const Point&, const int);
};
inline static void receiveSignalCreateBullet(...) { ... }
最后在文件C.cpp中,我使用signal_createBullet
这样:
signal_createBullet(bullet, pos, bulletVector, count);
A和B编译成功(使用g ++),但C失败并显示以下错误消息:
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>]
在尝试解决这个问题时,我将我的调用格式化,并在错误消息中第一个候选人更容易比较它们:
// 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'Portable'语法之外,(并且不,切换我的代码使用Portable风格没有区别)我看到两个调用之间没有区别,除了我调用的最后一件事是{ {1}}候选人的int&
。我尝试从信号中删除int
参数以查看是否存在问题,但事实并非如此。
任何人都明白为什么我会收到此错误?
答案 0 :(得分:1)
static void receiveSignalCreateBullet(const Bullet&amp;,const Point&amp;,const Point&amp;,const int);
为什么这里的参数是const?在信号声明中,它们不是常量。