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特定的C ++模板专业化

[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()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>operator()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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