Question

template <class ContainerType, class elementType>
void SerializeContainer (    ContainerType< elementType > container )
{
}

//call like this 
std::vector<int>  vector;
SerializeContainer(vector);

below will not compile.. Is there any way I can get through this?

Answer 1

It is easier to use the value_type member of the container to extract the element type:

  template <typename T>
  void SerializeContainer (const T& container)
  {
      typedef typename T::value_type ElementType;
      ..........
  }

---

The reason why your code does not work is because the syntax ContainerType<ElementType> only works when ContainerType is a really template. C++ support the template-template parameter for this:

template <template <typename> class ContainerType, typename ElementType>
void serializeContainer(const ContainerType<ElementType>& container) {
}

But even this will not match for standard C++ containers, because there are some hidden default arguments. When written explicitly:

std::vector<int, 
            std::allocate<int> >   vector;

So you’ll need to make the ContainerType parameter to accept 2 arguments:

template <typename <typename, typename> C, typename E, typename A>
void serializeContainer(const C<E, A>& container) {
}

But then this won’t match set::set because it has a different number of template parameters:

std::set<int, 
         std::less<int>, 
         std::allocator<int> >  set;

There simply isn’t a one-size-fit-all solution (before C++11) if you want the ElementType spelled explicitly when matching. It’s better to stick with the traits provided by the container.