C++11中std::forward的使用

std::forward argument: Returns an rvalue reference to arg if arg is not an lvalue reference; If arg is an lvalue reference, the function returns arg without modifying its type.

std::forward:This is a helper function to allow perfect forwarding of arguments taken as rvalue references to deduced types, preserving any potential move semantics involved. 

std::forward<T>(u)有两个参数：T 与 u。当T为左值引用类型时，u将被转换为T类型的左值，否则u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。

std::move是无条件的转为右值引用，而std::forward是有条件的转为右值引用，更准确的说叫做Perfect forwarding(完美转发)，而std::forward里面蕴含着的条件则是Reference Collapsing(引用折叠)。

std::move不move任何东西。std::forward也不转发任何东西。在运行时，他们什么都不做。不产生可执行代码，一个比特的代码也不产生。

std::move和std::forward只是执行转换的函数(确切的说应该是函数模板)。std::move无条件的将它的参数转换成一个右值，而std::forward当特定的条件满足时，才会执行它的转换。

std::move表现为无条件的右值转换，就其本身而已，它不会移动任何东西。 std::forward仅当参数被右值绑定时，才会把参数转换为右值。 std::move和std::forward在运行时不做任何事情。

下面是从其他文章中copy的测试代码，详细内容介绍可以参考对应的reference：

#include "forward.hpp"#include <utility>#include <iostream>#include <memory>#include <string>//////////////////////////////////////////////// reference: http://en.cppreference.com/w/cpp/utility/forwardstruct A {A(int&& n) { std::cout << "rvalue overload, n=" << n << "\n"; }A(int& n)  { std::cout << "lvalue overload, n=" << n << "\n"; }};class B {public:template<class T1, class T2, class T3>B(T1&& t1, T2&& t2, T3&& t3) :a1_{ std::forward<T1>(t1) },a2_{ std::forward<T2>(t2) },a3_{ std::forward<T3>(t3) }{}private:A a1_, a2_, a3_;};template<class T, class U>std::unique_ptr<T> make_unique1(U&& u){return std::unique_ptr<T>(new T(std::forward<U>(u)));}template<class T, class... U>std::unique_ptr<T> make_unique(U&&... u){return std::unique_ptr<T>(new T(std::forward<U>(u)...));}int test_forward1(){auto p1 = make_unique1<A>(2); // rvalueint i = 1;auto p2 = make_unique1<A>(i); // lvaluestd::cout << "B\n";auto t = make_unique<B>(2, i, 3);return 0;}////////////////////////////////////////////////////////// reference: http://www.cplusplus.com/reference/utility/forward/// 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 lvalueoverloaded(std::forward<T>(x));  // rvalue if argument is rvalue}int test_forward2(){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;}//////////////////////////////////////////////////////// reference: http://stackoverflow.com/questions/8526598/how-does-stdforward-worktemplate<class T>struct some_struct{T _v;template<class U>some_struct(U&& v) : _v(static_cast<U&&>(v)) {} // perfect forwarding here// std::forward is just syntactic sugar for this};int test_forward3(){/* remember the reference collapsing rules(引用折叠规则):前者代表接受类型，后者代表进入类型，=>表示引用折叠之后的类型，即最后被推导决断的类型TR   RT&   &->T&   // lvalue reference to cv TR -> lvalue reference to TT&   &&->T&  // rvalue reference to cv TR -> TR (lvalue reference to T)T&&  &->T&   // lvalue reference to cv TR -> lvalue reference to TT&&  &&->T&& // rvalue reference to cv TR -> TR (rvalue reference to T) */some_struct<int> s1(5);// in ctor: '5' is rvalue (int&&), so 'U' is deduced as 'int', giving 'int&&'// ctor after deduction: 'some_struct(int&& v)' ('U' == 'int')// with rvalue reference 'v' bound to rvalue '5'// now we 'static_cast' 'v' to 'U&&', giving 'static_cast<int&&>(v)'// this just turns 'v' back into an rvalue// (named rvalue references, 'v' in this case, are lvalues)// huzzah, we forwarded an rvalue to the constructor of '_v'!// attention, real magic happens hereint i = 5;some_struct<int> s2(i);// in ctor: 'i' is an lvalue ('int&'), so 'U' is deduced as 'int&', giving 'int& &&'// applying the reference collapsing rules yields 'int&' (& + && -> &)// ctor after deduction and collapsing: 'some_struct(int& v)' ('U' == 'int&')// with lvalue reference 'v' bound to lvalue 'i'// now we 'static_cast' 'v' to 'U&&', giving 'static_cast<int& &&>(v)'// after collapsing rules: 'static_cast<int&>(v)'// this is a no-op, 'v' is already 'int&'// huzzah, we forwarded an lvalue to the constructor of '_v'!return 0;}////////////////////////////////////////////////////// reference: https://oopscenities.net/2014/02/01/c11-perfect-forwarding/void sum(int a, int b){std::cout << a + b << std::endl;}void concat(const std::string& a, const std::string& b){std::cout<< a + b << std::endl;}void successor(int a, int& b){b = ++a;}template <typename PROC, typename A, typename B>void invoke(PROC p, A&& a, B&& b){p(std::forward<A>(a), std::forward<B>(b));}int test_forward4(){invoke(sum, 10, 20);invoke(concat, "Hello", "world");int s = 0;invoke(successor, 10, s);std::cout << s << std::endl;return 0;}

GitHub：https://github.com/fengbingchun/Messy_Test