Perfect forwarding with variadic template function

Peter87 (11254)
I think what I'm trying to do is called perfect forwarding but I can't get it to work.
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#include <iostream>

void bar(int&) {std::cout << "int&" << std::endl;}
void bar(const int&) {std::cout << "const int&" << std::endl;}
void bar(int&&) {std::cout << "int&&" << std::endl;}

template <typename... Args>
void foo(Args&&... args)
{
	return bar(args...);
}

int main()
{
	int i = 0;
	const int ci = 0;
	
	foo(i);
	foo(ci);
	foo(5);
}
 
int&
const in&
int&


I want the result to be the same as when I call bar directly.
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bar(i);
bar(ci);
bar(5);
 
int&
const in&
int&&


What am I doing wrong?
Cubbi (4774)
Inside your void foo(), each arg is an lvalue. You need to use std::forward to turn it back into rvalue if it needs to be one

http://en.cppreference.com/w/cpp/utility/forward

example: http://ideone.com/AxBD0
Last edited on
Peter87 (11254)
Thank you that works perfect.
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