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時在文件A.hpp,我有「不匹配調用」 編譯器錯誤使用boost ::信號
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,我有一個類Entities
具有靜態成員函數receiveSignalCreateBullet
,我想與signal_createBullet
連接,像這樣:(命名空間不再贅述)
class Entities
{
Entities()
{
signal_createBullet.connect(&receiveSignalCreateBullet);
}
public:
static void receiveSignalCreateBullet(const Bullet&, const Point&, const Point&, const int);
};
inline static void receiveSignalCreateBullet(...) { ... }
終於在文件C.cpp,我使用signal_createBullet
這樣的:
signal_createBullet(bullet, pos, bulletVector, count);
甲乙編譯成功(使用克++),但是C失敗,此錯誤消息:
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>]
雖然試圖弄清楚這一點,我格式化我的電話,並在錯誤消息的第一候選人更容易對它們進行比較:
// 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
除了候選人使用'便攜'語法的事實(不,切換我的代碼使用便攜式風格沒有區別)我看到兩個電話之間沒有區別,除了我呼叫的最後一個東西是int&
候選人有一個int
。我嘗試從信號中去除int
參數,看看這是否是問題,而事實並非如此。
任何人都明白爲什麼我會收到此錯誤?
啊!他們最初是const的,雖然我在誇耀我的代碼試圖讓事情發揮作用,但是我讓它們不是const,並且忘記將它們改回來。感謝您爲我偵察明顯。 :) – Max 2011-05-11 22:30:01