Reputation: 2564
I am trying to understand the concept of forwarding in C++ and wrote below code to understand this functionality on variadic templates.
#include<utility>
#include<string>
#include<tuple>
#include<sstream>
template<typename... Args> struct Helper_Class
{
std::tuple<Args...> argTuple;
Helper_Class(Args&&... args):
argTuple(std::make_tuple(std::forward<Args>(args)...))
{}
};
template<typename... Args> std::ostream&
operator<< ( std::ostream& os,Helper_Class<Args...> obj)
{
return os;
}
template<typename...Args>
Helper_Class< Args...>
Print(Args&&... args)
{
return Helper_Class<Args...>(std::forward<Args>(args)...);
}
template <typename... Ts>
void test( Ts &&...params) {
std::stringstream s;
s <<Print(std::forward<Ts>(params)...);
}
int main()
{
test(1,2,"foo", 'x');
}
However, while executing the code I am getting below error:
error: no matching function for call to ‘std::tuple<int, int, const char (&)[4], char>::tuple(std::tuple<int, int, const char*, char>)’
argTuple(std::make_tuple(std::forward<Args>(args)...))
What I am not understanding is if I am trying to do perfect forwarding (means types should not change between function calls) why is it trying to change the type?
Please excuse if this is very basic as I am learning this concept for very first time.
Update 1: Continuation question posted on:Question
Upvotes: 3
Views: 747
Reputation: 118330
template <typename... Ts>
void test( Ts &&...params) {
std::stringstream s;
s <<Print(std::forward<Ts>(params)...);
}
You're going to start here. The template parameters to Print
will come out of std::forward
. As such, they will already be forwarded lvalue/rvalue references, et al. That's what you are going to shove here, as parameters to the Print
template.
Then:
template<typename...Args>
Helper_Class< Args...>
Print(Args&&... args)
This is now going to forward Args...
to Helper_Class
, which will use those types to try to declare a tuple containing lvalue references, rvalue references, et al. That's not going to work very well.
You need to decay these suckers:
template<typename...Args>
Helper_Class< typename std::decay<Args>::type...>
Print(Args&&... args)
{
return Helper_Class<typename std::decay<Args>::type...>(std::forward<Args>(args)...);
}
Or, let's use C++14 to clean up some visual clutter...
template<typename...Args>
auto Print(Args&&... args)
{
return Helper_Class<typename std::decay<Args>::type...>(std::forward<Args>(args)...);
}
This now compiled for me, with gcc 6.2.1.
P.S. This is not exactly "basic"...
Upvotes: 5