传递模板化智能指针类型作为模板参数

Question

I am trying to pass smart_pointer as a template parameter than use the type for dynamic casting etc..

But there is one more catch smart_pointer can have many types like std::shared_ptr, boost::shared_ptr, std::weak_ptr so on.

So what I am trying finally is replace typename SmartPointerType<DerivedClass> with correct syntax :

template < typename SmartPointerType<DerivedClass> >
std::vector<SmartPointerType<DerivedClass>> filterByType()
{
    std::vector<SmartPointerType<DerivedClass>> newList;
    for ( int i = 0; i < iTerrainList.size(); i++)
    {
        SmartPointerType<DerivedClass> castedTerrain = Cast<DerivedClass>(iTerrainList[i]);
        if ( castedTerrain )
            newList.push_back(castedTerrain);
    }

    return newList;
}

"Cast<>()" accepts any type of smart pointer that I am interested btw.

I find out I can use something like

template <template <typename Type> class TemplateType>

But it didn't worked for with my c++03 compiler.

I know I can use ".get()" but I am curious how can I make it work without ".get()".

Answer 1

Just let it work on any type:

template <typename DerivedClassPtr>
std::vector<DerivedClassPtr> filterByType()
{
    std::vector<DerivedClassPtr> newList;
    for ( int i = 0; i < iTerrainList.size(); i++)
        if ( DerivedClassPtr terrain = Cast<typename DerivedClassPtr::element_type>(iTerrainList[i]) )
            newList.push_back(terrain);

    return newList;
}

Here, the requirements on the template type are that it:

• Can be stored in a vector (so must be copyable).
• Has element_type (so raw pointers won't work, for better or worse; you could fix this with type traits implemented yourself in C++03).
• Is accepted by your Cast() function.
• Is usable in a boolean context.