我为什么要使用std :: forward？

In the code below, why should I use std::forward when passing around arguments?

在下面的代码中,为什么我在传递参数时使用std :: forward?

class Test {
  public:
  Test() {
      std::cout << "ctor" << std::endl;
  }
  Test(const Test&) {
      std::cout << "copy ctor" << std::endl;
  }
  Test(const Test&&) {
      std::cout << "move ctor" << std::endl;
  }
};

template<typename Arg>
void pass(Arg&& arg) {
    // use arg..
    return; 
}

template<typename Arg, typename ...Args>
void pass(Arg&& arg, Args&&... args)
{
    // use arg...
    return pass(args...); // why should I use std::forward<Arg>(args)... ?
}

int main(int argc, char** argv)
{
    pass(std::move<Test>(Test()));

    return 0;
}

The code with or without std::forward doesn't show any copy/move around.

带或不带std :: forward的代码不显示任何复制/移动。

1 个解决方案

#1

There are a number of good posts on what std::forward does and how it works (such as here and here).

关于std :: forward做什么以及如何工作(例如这里和这里)有很多好的帖子。

In a nutshell, it preserves the value category of its argument. Perfect forwarding is there to ensure that the argument provided to a function is forwarded to another function (or used within the function) with the same value category (basically r-value vs. l-value) as originally provided. It is typically used with template functions where reference collapsing may have taken place (involving universal/forwarding references).

简而言之,它保留了其参数的值类别。完美转发是为了确保提供给函数的参数被转发到另一个函数(或在函数内使用),其具有与最初提供的相同的值类别(基本上是r值与l值)。它通常与模板函数一起使用,其中可能发生了引用折叠(涉及通用/转发引用)。

Consider the code sample below. Removing the std::forward would print out requires lvalue and adding the std::forward prints out requires rvalue. The func is overloaded based on whether it is an rvalue or an lvalue. Calling it without the std::forward calls the incorrect overload. The std::forward is required in this case as pass is called with an rvalue.

考虑下面的代码示例。删除std :: forward将打印输出需要左值并添加std :: forward打印输出需要rvalue。根据fval是rvalue还是左值,func会被重载。在没有std :: forward的情况下调用它会调用不正确的重载。在这种情况下需要std :: forward,因为使用rvalue调用pass。

#include <utility>
#include <iostream>
class Test {
  public:
  Test() {
      std::cout << "ctor" << std::endl;
  }
  Test(const Test&) {
      std::cout << "copy ctor" << std::endl;
  }
  Test(Test&&) {
      std::cout << "move ctor" << std::endl;
  }
};

void func(Test const&)
{
    std::cout << "requires lvalue" << std::endl;
}

void func(Test&&)
{
    std::cout << "requires rvalue" << std::endl;
}

template<typename Arg>
void pass(Arg&& arg) {
    // use arg here
    func(std::forward<Arg>(arg));
    return; 
}

template<typename Arg, typename ...Args>
void pass(Arg&& arg, Args&&... args)
{
    // use arg here
    return pass(std::forward<Args>(args)...);
}

int main(int, char**)
{
    pass(std::move<Test>(Test()));
    return 0;
}