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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?
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