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,它有一个我想连接Entities
的静态成员函数,如下所示:(为简洁起见,省略了命名空间)receiveSignalCreateBullet
signal_createBullet
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signal_createBullet
这样使用:
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>]
在试图弄清楚这一点时,我格式化了我的调用和错误消息中的第一个候选者,以便更轻松地比较它们:
// 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“便携式”语法这一事实(不,将我的代码切换为使用便携式风格没有区别)我认为这两个调用之间没有区别,除了我调用中的最后一件事是int&
候选人有一个int
。我尝试int
从信号中删除参数以查看是否是问题所在,但事实并非如此。
有人知道为什么我会收到此错误吗?