move forward

C++11 STL

move

template <class T>
typename remove_reference<T>::type&& move (T&& arg) noexcept;

　　我对move的理解就是，它使得程序员在拷贝，赋值时更好利用一次性的局部变量。
Returns an rvalue reference to arg.
The function returns the same as:static_cast<remove_reference<decltype(arg)>::type&&>(arg)
example

#include <utility>      // std::move
#include <iostream>     // std::cout
#include <vector>       // std::vector
#include <string>       // std::string
int main () {
  std::string foo = "foo-string";
  std::string bar = "bar-string";
  std::vector<std::string> myvector;
  myvector.push_back (foo);                    // copies
  myvector.push_back (std::move(bar));         // moves
  std::cout << "myvector contains:";
  for (std::string& x:myvector) std::cout << ' ' << x;
  std::cout << '\n';
  return 0;
}

forward

template <class T> T&& forward (typename remove_reference<T>::type& arg) noexcept;
template <class T> T&& forward (typename remove_reference<T>::type&& arg) noexcept;

函数功能:当T为左值引用类型时，arg将被转换为T类型的左值，否则arg将被转换为T类型右值。如此定义std::forward是为了在使用右值引用参数的函数模板中解决参数的完美转发问题。完美转发是指保留参数的左右值属性，从而实现了函数模板参数的完美转发。如下图，在fn中，x已经是具名变量，变成了左值了，但是在fn被调用时，x是属于右值变量，而forward就是继续保持它的右值属性。
example

#include <utility>      // std::forward
#include <iostream>     // std::cout
// function with lvalue and rvalue reference overloads:
void overloaded (const int& x) {std::cout << "[lvalue]";}
void overloaded (int&& x) {std::cout << "[rvalue]";}
// function template taking rvalue reference to deduced type:
template <class T> void fn (T&& x) {
  overloaded (x);                   // always an lvalue
  overloaded (std::forward<T>(x));  // rvalue if argument is rvalue
}
int main () {
  int a;
  std::cout << "calling fn with lvalue: ";
  fn (a);
  std::cout << '\n';
  std::cout << "calling fn with rvalue: ";
  fn (0);
  std::cout << '\n';
  return 0;
}