Question

I have a template function that I wish to return either the type of T or a variant. I tried to do as follows, however the compiler complains it cannot convert ‘variant’ to int (where I use this function with T=int).

How should I implement this so I can either just return the variant or the type contain in the variant.

It is gotten out of a vector structs.

template <typename T>
T find_attribute(const std::string& attribute, bool isVariant = false)
{
    std::vector<boost::shared_ptr<node> >::iterator nodes_iter = _request->begin();

    for (; nodes_iter != _request->end(); nodes_iter++)
    {
        size_t sz = (*nodes_iter)->attributes.size();
        std::vector<node::attrib>::iterator att_iter = (*nodes_iter)->attributes.begin();
        for (; att_iter != (*nodes_iter)->attributes.end(); att_iter++)
        {
            if (att_iter->key.compare(attribute) == 0)
            {
                if (isVariant)
                {
                    return att_iter->value; //return variant
                }
                else
                {
                    return boost::get<T>(att_iter->value); // return type inside variant as given by T.
                }
            }
        }
    }
}

Answer 1

You can create a template specialisation for find_attribute<boost::variant>(const std::string& attribute) that return a variant and a normal version attribute<T>(const std::string& attribute).

The normal version would do:

return boost::get<T>(find_attribute<variant>(attribute));

But remeber that template are evaluated at compile time!

If find_attribute is a member function, you can use this only with the msvc compiler.

If you can’t do template specialisation, you could name the functions different.