C++ perfect forwarding

When you want to forward arguments to template function/method without changing the value category

Perfect forwarding

refers to a template function or method which passes arguments

• 1. to another function or method
• 2. preserving the const qualifier and lvalue/rvalue category

std::move( ) vs std::forward( )

std::move

• force the value category as a rvalue. It is basically a casting to rvalue

std::forward

• pass in an original reference type. If what is passed is a rvalue, then convert it as a rvalue category. If what is passed to is a lvalue, then pass as a lvalue category.

#include <iostream>
#include <vector>
using namespace std;
class Foo
{
public:
  Foo() = default;
  Foo(int a, bool b, const vector<int> &v)
      : mA(a), mB(b), mV(v) {}
  Foo(int a, bool b, vector<int> &&v)
      : mA(a), mB(b), mV(move(v)) {}
  virtual ~Foo() {}
private:
  int mA;
  bool mB;
  vector<int> mV;
};
class Bar
{
public:
  template <typename... Args>
  void addFoo(Args &&...args)
  {
    mFoo.emplace_back(forward<Args>(args)...);
  }
private:
  vector<Foo> mFoo;
};
static void printVector(vector<int> const &v)
{
  for (auto e : v)
  {
    cout << e << endl;
  }
}
int main()
{
  Bar b;
  vector<int> v1 = {1, 2, 3, 4, 5};
  vector<int> v2 = {6, 7, 8, 9};
  b.addFoo(1, true, v1);        // Foo will receive it as lvalue ref
  b.addFoo(2, false, move(v2)); // Foo will receive it as rvalue ref
  cout << "V1 ------" << endl;
  printVector(v1);
  cout << "V2 ------" << endl;
  printVector(v2);
  return 0;
}

result

~/ProgramingPractice/C++$ ./a.out 
V1 ------
1
2
3
4
5
V2 ------