C++11新特性之 std::forward(完美转发)

我们也要时刻清醒，有时候右值会转为左值，左值会转为右值。 
（也许“转换”二字用的不是很准确）

如果我们要避免这种转换呢？ 
我们需要一种方法能按照参数原来的类型转发到另一个函数中，这才完美，我们称之为完美转发。

std::forward就可以保存参数的左值或右值特性。

因为是这样描述的： 
When used according to the following recipe in a function template, forwards the argument to another function with the value category it had when passed to the calling function.

例子：

template<class T>
void wrapper(T&& arg)
{
    foo(std::forward<T>(arg)); // Forward a single argument.
}

template<class T>
void wrapper(T&& arg)
{
    foo(std::forward<T>(arg)); // Forward a single argument.
}

If a call to wrapper() passes an rvalue std::string, then T is deduced to std::string (not std::string&, const std::string&, or std::string&&), and std::forward ensures that an rvalue reference is passed to foo. 
If a call to wrapper() passes a const lvalue std::string, then T is deduced to const std::string&, and std::forward ensures that a const lvalue reference is passed to foo. 
If a call to wrapper() passes a non-const lvalue std::string, then T is deduced to std::string&, and std::forward ensures that a non-const lvalue reference is passed to foo.

看一段网站上的代码(http://en.cppreference.com/w/cpp/utility/forward)：

#include <iostream>
#include <memory>
#include <utility>
#include <array>

struct 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 main()
{
    auto p1 = make_unique1<A>(); // rvalue
    int i = ;
    auto p2 = make_unique1<A>(i); // lvalue

    std::cout << "B\n";
    auto t = make_unique<B>(, i, );
}
//输出：
rvalue overload, n=
lvalue overload, n=
B
rvalue overload, n=
lvalue overload, n=
rvalue overload, n=

最后，记住： 
不管是T&&、左值引用、右值引用，std::forward都会按照原来的类型完美转发。

可能还不是很清楚,再举个栗子

#include <iostream>
using namespace std;

void F(int a) {
    cout << a << endl;
}

void F(int&& a) {　　　　//int&&并不是右值了，只是它能被右值初始化,记住右值引用a是一个绑定了右值对象的左值
    // do something
}

template<class A>
void G(A &&a) {
    return F(std::forward<A>(a));  //
    return F(a);        //
}

int main() {
    int i = ;
    G(i);
    G();

    system("pause");
}

例子中,如果你不使用std::forward转发，G中将始终调用void F(int)这个版本,即使G的参数是个右值也不会调用void F(int&& a)