C ++转发参考和R值参考 [英] C++ forwarding reference and r-value reference

问题描述

我知道转发引用是对cv不合格模板参数的右值引用"，例如

I understand that a forwarding reference is "an rvalue reference to a cv-unqualified template parameter", such as in

template <class T> void foo(T&& );

这意味着上面的函数可以同时使用l值和r值引用.

which means the above function can take both l-value and r-value reference.

有些事情我不理解，例如

There's something I don't understand, e.g.

template <class T>
class A
{
    template <class U>
    void foo(T&& t, U&& u)
    {
        T t2( std::forward(t) ); // or should it be std::move(t)? is T&& forwarding or r-value reference
        U u2( std::forward(u) ); // or should it be std::move(u)? I believe U&& is forwarding reference
    }
};

在上面的代码中，都是T&&和U&&转发参考?

我写了一些代码进行测试(VS2015编译器):

I wrote some code to test (VS2015 compiler):

class A
{
public:
    A(){};
    A(const A& rhs)
    {
        std::cout << "calling 'const A&' l-value" << std::endl;
    }

    A(A&& rhs)
    {
        std::cout << "calling ' A&&' r-value" << std::endl;
    }

};

template <class T>
class Test
{
public:
    void test1(T&& t)
    {
        T t2(std::forward<T>(t));
    }

    template <typename X>
    void test2(X&& x)
    {
        T t2( std::forward<T>( x ) );
    }

};

void main()
{
    A a;
    Test<A> test;
    test.test1(A());
    test.test1(std::move(a));
    //test.test1(a); // this doesn't compile. error: cannot convert argument 1 from 'A' to 'A &&', You cannot bind an lvalue to an rvalue reference

    test.test2<A>(A());
    test.test2<A>( std::move( a ) );

    //test.test2<A>( a ); // this doesn't compile. error: cannot convert argument 1 from 'A' to 'A &&', You cannot bind an lvalue to an rvalue reference
}

我期待的是test.test1(a);如果test.test2(a)正在转发引用，则两者都应编译，但两者都不应该.

I was expecting that test.test1(a); and test.test2(a) should both compile if they are forwarding references, but neither does.

有人可以向我解释吗?谢谢！

Could someone explain this to me? Thanks!

修改- - - - - - - 谢谢你们 - - - - - -理查德(Richard)和阿尔蒂(Artemy)是正确的.

Edit --------------thanks, guys----------- Richard and Artemy are correct.

推荐答案

这是一个很好的问题，一开始几乎困扰所有人.

It's a great question which foxes almost everyone in the beginning.

template <class T>
class A
{
    template <class U>
    void foo(T&& t, U&& u);
};

在此示例中，未推导 T (您在实例化模板时明确定义了它).

In this example, T is not deduced (you explicitly define it when you instanciate the template).

U 是因为它是从参数 u 推导出来的.

U is deduced because it's deduced from the argument u.

因此，几乎在所有情况下都是:

Therefore, in almost all cases it would be:

std::move(t);
std::forward<U>(u);

这篇关于C ++转发参考和R值参考的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！