[英]Specific C++ template specialization
I have the following code fragment: 我有以下代码片段:
template <class T>
struct ServicePtr
{
std::shared_ptr<T> service;
~ServicePtr()
{
std::cout << __PRETTY_FUNCTION__ << std::endl;
}
};
template <class T>
struct ServicePtrDeleter
{
void operator()(ServicePtr<T>* ref) const
{
if (ref->service.get())
{
if (IRunnable<T>* r=dynamic_cast<IRunnable<T>*>(ref->service.get()))
{
//std::cout << "Mark Thread Finished" << std::endl;
r->stop();
}
else
{
std::cout << __FILE__ << ":" << __LINE__ << std::endl;
}
}
delete ref;
}
};
template <typename T>
struct ServiceCreator
{
static std::shared_ptr< ServicePtr<T> > create()
{
std::shared_ptr< ServicePtr<T> > servicePtr(new ServicePtr<T>, ServicePtrDeleter< ServicePtr<T> >());
servicePtr->service.reset(new T);
if (IRunnable<T>* r=dynamic_cast<IRunnable<T>*>(servicePtr->service.get()))
{
r->setSelfPtr(std::dynamic_pointer_cast<IRunnable<T>>(servicePtr->service));
}
else
{
std::cout << __FILE__ << ":" << __LINE__ << std::endl;
}
return servicePtr;
}
};
ServiceCreator<T>
, with T may or may not be derivated from IRunnable. 带有T的
ServiceCreator<T>
可以或可以不从IRunnable派生。 I get the following compiler error(GCC 4.6.1): 我收到以下编译器错误(GCC 4.6.1):
In file included from /usr/lib/gcc/i686-pc-linux-gnu/4.6.1/../.. /../../include/c++/4.6.1/bits/shared_ptr.h:52:0,
from /usr/lib/gcc/i686-pc-linux-gnu/4.6.1/../../../../include/c++/4.6.1/memory:86,
from /usr/lib/gcc/i686-pc-linux-gnu/4.6.1/../../../../include/c++/4.6.1/thread:40,
from main.cc:2:
/usr/lib/gcc/i686-pc-linux-gnu/4.6.1/../../../../include/c++/4.6.1/bits/shared_ptr_base.h: In constructor 'std::__shared_count<_Lp>::__shared_count(_Ptr, _Deleter) [with _Ptr = ServicePtr<LogWriter>*, _Deleter = ServicePtrDeleter<ServicePtr<LogWriter> >, __gnu_cxx::_Lock_policy _Lp = (__gnu_cxx::_Lock_policy)2u]':
/usr/lib/gcc/i686-pc-linux-gnu/4.6.1/../../../../include/c++/4.6.1/bits/shared_ptr_base.h:771:37: instantiated from 'std::__shared_ptr<_Tp, _Lp>::__shared_ptr(_Tp1*, _Deleter) [with _Tp1 = ServicePtr<LogWriter>, _Deleter = ServicePtrDeleter<ServicePtr<LogWriter> >, _Tp = ServicePtr<LogWriter>, __gnu_cxx::_Lock_policy _Lp = (__gnu_cxx::_Lock_policy)2u]'
/usr/lib/gcc/i686-pc-linux-gnu/4.6.1/../../../../include/c++/4.6.1/bits/shared_ptr.h:128:37: instantiated from 'std::shared_ptr<_Tp>::shared_ptr(_Tp1*, _Deleter) [with _Tp1 = ServicePtr<LogWriter>, _Deleter = ServicePtrDeleter<ServicePtr<LogWriter> >, _Tp = ServicePtr<LogWriter>]'
ServiceCreator.h:45:102: instantiated from 'static std::shared_ptr<ServicePtr<U> > ServiceCreator<T>::create() [with T = LogWriter]'
main.cc:114:27: instantiated from here
/usr/lib/gcc/i686-pc-linux-gnu/4.6.1/../../../../include/c++/4.6.1/bits/shared_ptr_base.h:480:8: error: no match for call to '(ServicePtrDeleter<ServicePtr<LogWriter> >) (ServicePtr<LogWriter>*&)'
ServiceCreator.h:19:8: note: candidate is:
ServiceCreator.h:21:7: note: void ServicePtrDeleter<T>::operator()(ServicePtr<T>*) const [with T = ServicePtr<LogWriter>]
ServiceCreator.h:21:7: note: no known conversion for argument 1 from 'ServicePtr<LogWriter>*' to 'ServicePtr<ServicePtr<LogWriter> >*'
Why ServicePtrDeleter::operator()
gets instantiated with ServicePtr< ServicePtr<T> >
? 为什么
ServicePtrDeleter::operator()
用ServicePtr< ServicePtr<T> >
实例化? I just want to get T, inside operator()
, so I can test whether it implements IRunnable
or not. 我只想在
operator()
内获取T,所以我可以测试它是否实现IRunnable
。
main.cc:114 is auto logWriter=LogWriter::create();
main.cc:114是
auto logWriter=LogWriter::create();
It is a global variable. 它是一个全局变量。
class LogWriter:
public ServiceCreator<LogWriter>,
public IRunnable<LogWriter>,
{
....
}
ServicePtrDeleter
is instantiated here: 在此处实例化
ServicePtrDeleter
:
std::shared_ptr< ServicePtr<T> > servicePtr(new ServicePtr<T>, ServicePtrDeleter< ServicePtr<T> >());
The template parameter given to ServicePtrDeleter
is ServicePtr<T>
. 提供给
ServicePtrDeleter
的模板参数是ServicePtr<T>
。
When you replace the template parameter with ServicePtr<T>
in operator()
of ServicePtrDeleter
, you end up with a ServicePtr< ServicePtr<T> >
parameter. 当更换与模板参数
ServicePtr<T>
在operator()
的ServicePtrDeleter
,则结束了一个ServicePtr< ServicePtr<T> >
参数。
You probably just wanted to use a T
template parameter in the instantiation, since ServicePtrDeleter
wraps that into a ServicePtr
itself: 您可能只想在实例化中使用
T
模板参数,因为ServicePtrDeleter
将其包装到ServicePtr
本身中:
... servicePtr(new ServicePtr<T>, ServicePtrDeleter<T>());
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