C++11 Perfect forwarding question

[o3hC5Di1]

Hi everyone,

I'm slowly starting to understand rvalue references and their benefits.
At the moment I'm creating a templated facade class, so I thought it might be the perfect time to use perfect forwarding. Unfortunately the compiler, gcc 4.6.3, is throwing me this error:

no known conversion for argument 1 from ‘type’ to ‘type&&’

Simplified version of the code I'm using:

template <typename EntityType, typename EntityMapper>
class facade
{
    public:
        void read(EntityType&& e)
        {
            _mapper.read(std::forward<EntityType>(e));
        }

    private:
        EntityMapper _mapper;
};

//used:
facade<person, personmapper> personfacade;
person p;
personfacade.read(p);  //this fails
personfacade.read(std::move(p)); //this works 

Could anyone please explain the difference, as to why this doesn't work, but the factory function on this page does: http://thbecker.net/articles/rvalue_references/section_08.html

To me they seem to be pretty much the same, but I must be missing something.
I'm trying to understand this, not to just copy things.

Thanks very much in advance.

All the best,
NwN

[Catfish3]

Isn't p an lvalue in your code?
personfacade.read(person());

[Cubbi]

read needs to be a template function (that is,

template<typename T>
void read(T&& e)

in order to bind to both lvalues and rvalues (in which case std::forward can be used to forward the value category elsewhere)

[o3hC5Di1]

Catfish3 wrote:

Isn't p an lvalue in your code?

Yes... but I understood the link I included in my post as "if you define a templated function with Type&&, the compiler will create versions as needed for both lvalues and rvalues".

I'm obviously missing something?

All the best,
NwN

[o3hC5Di1]

Cubbi wrote:

read needs to be a template function

Hmm, I sort of hoped it would suffice to make it a member function of a class template.
Is there any alternative?

Thanks very much for your help guys :)

All the best,
NwN

[Cubbi]

Just like that

    public:
        template<typename T>
        void read(T&& e)
        {
            _mapper.read(std::forward<T>(e));
        }

[o3hC5Di1]

Then wouldn't I have to do following?

facade<person, personmapper> personfacade;
person p;
personfacade.read<person>(p); 

-N

[Cubbi]

No, that would not compile. personfacade.read(p); is the intended use case.

[o3hC5Di1]

That's exactly what I was after, thank you very much!
I think I can figure out whatever questions I have from the page I referred to.

Thanks again.

All the best,
NwN