template <class... ArgTypes>
typename result_of<T&(ArgTypes&&... )>::type
operator()(ArgTypes&&... args) const;
Returns: INVOKE(get(), std::forward<ArgTypes>(args)...). ([func.require])
Remark: operator() is described for exposition only. Implementations are not required to provide an actual reference_wrapper::operator(). Implementations are permitted to support reference_wrapper function invocation through multiple overloaded operators or through other means.