接受右值和左值引用的可变参数模板？

Question

我想编写一个接受右值和左值引用的可变参数模板函数。它将大写 std::strings，并在大写后显示每个参数。函数结束后，所有左值都应保持大写（即通过引用传递左值）。
例如，我想要这种行为：

std::string hello = "hello";
std::string planet = "planet";
std::string earth = "earth";

//the order and amount of rvalues and lvalue references, should not matter
Capitalize_And_Output("hello","planet","earth"); //outputs: "HELLO PLANET EARTH"
Capitalize_And_Output(hello,"planet","earth");   //outputs: "HELLO PLANET EARTH"
Capitalize_And_Output("hello",planet,"earth");   //outputs: "HELLO PLANET EARTH"
Capitalize_And_Output("hello","planet",earth);   //outputs: "HELLO PLANET EARTH"
Capitalize_And_Output(hello,planet,"earth");     //outputs: "HELLO PLANET EARTH"
Capitalize_And_Output(hello,"planet",earth);     //outputs: "HELLO PLANET EARTH"
Capitalize_And_Output("hello",planet,earth);     //outputs: "HELLO PLANET EARTH"
Capitalize_And_Output(hello,planet,earth);       //outputs: "HELLO PLANET EARTH"

//lvalue references remain changed after the function call
std::cout << hello << std::endl;  //outputs: "HELLO"
std::cout << planet << std::endl; //outputs: "PLANET"
std::cout << earth << std::endl;  //outputs: "WORLD"

我怎样才能得到上面的代码块来编译和工作，如图所示？

到目前为止，我能够输出信息，但我不知道如何处理两种不同值类型的大写。以下代码将编译，因为我已经注释掉了不起作用的行。

#include <string>
#include <iostream>
#include <algorithm>

template<typename T>
void Capitalize_And_Output(T & str) {
    //std::transform(str.begin(), str.end(), str.begin(), ::toupper); <- will not compile
    std::cout << str<< std::endl;
    return;
}

template<typename First, typename ... Strings>
void Capitalize_And_Output(First & str, const Strings&... rest) {
    //std::transform(str.begin(), str.end(), str.begin(), ::toupper); <- will not compile
    std::cout << str << " ";
    Capitalize_And_Output(rest...);
    return;
}

int main() {

    std::string hello = "hello";
    std::string planet = "planet";
    std::string earth = "earth";

    //the order and amount of rvalues and lvalue references, should not matter
    Capitalize_And_Output("hello","planet","earth"); //outputs: "HELLO PLANET EARTH"
    Capitalize_And_Output(hello,"planet","earth");   //outputs: "HELLO PLANET EARTH"
    Capitalize_And_Output("hello",planet,"earth");   //outputs: "HELLO PLANET EARTH"
    Capitalize_And_Output("hello","planet",earth);   //outputs: "HELLO PLANET EARTH"
    Capitalize_And_Output(hello,planet,"earth");     //outputs: "HELLO PLANET EARTH"
    Capitalize_And_Output(hello,"planet",earth);     //outputs: "HELLO PLANET EARTH"
    Capitalize_And_Output("hello",planet,earth);     //outputs: "HELLO PLANET EARTH"
    Capitalize_And_Output(hello,planet,earth);       //outputs: "HELLO PLANET EARTH"

    //lvalue references keep changed value after the function call
    std::cout << hello << std::endl;  //outputs: "HELLO"
    std::cout << planet << std::endl; //outputs: "PLANET"
    std::cout << earth << std::endl;  //outputs: "WORLD"

    return 0;
}

我怎样才能让它工作？
也许转换函数不起作用，因为右值实际上是不同的类型？它们是 char*s 吗？

我在想什么：
我必须对类型特征做些什么吗？
有 R 值参考的东西？
具有普遍敬意的东西（不真的确定那是什么）？

请更正任何滥用术语的地方！

Answer 1

首先，您的程序无法运行的根本原因不仅与右值/左值兼容性有关。为了表明考虑这个版本只通过值传递参数

#include <string>
#include <iostream>
#include <algorithm>

template<typename T>
void Capitalize_And_Output(T str) {
    std::transform(str.begin(), str.end(), str.begin(), ::toupper); 
    std::cout << str<< std::endl;
    return;
}

template<typename First, typename ... Strings>
void Capitalize_And_Output(First str, Strings... rest) {
    std::transform(str.begin(), str.end(), str.begin(), ::toupper); 
    std::cout << str << " ";
    Capitalize_And_Output(rest...);
    return;
}

在这种情况下，您的主代码也将不起作用。为了表明这一点，请考虑以下内容

int main() {    
    std::string hello = "hello";
    std::string planet = "planet";
    std::string earth = "earth";
    Capitalize_And_Output(std::string("hello"),planet,earth);  // this will work
    //Capitalize_And_Output("hello",planet,earth);  // original code -> this will NOT work
    return 0;
}

线

Capitalize_And_Output("hello",planet,earth);

不起作用，因为编译器考虑第一个参数“a const char”（char 没有 .begin() 和 .end() 迭代器！）！。事实上，我在g++中收到以下错误消息

instantiated from here
teste.cpp:14:5: error: request for member ‘begin’ in ‘str’, which is of non-class type ‘const char*’

现在回到你原来的左值/右值问题（折叠规则说 && & = &。这是解决方案。你也不能忘记“完美转发”右值引用。

那么，总结一下最后的答案就是

template<typename T>
void Capitalize_And_Output(T&& str) {
    std::transform(str.begin(), str.end(), str.begin(), ::toupper); // <- will not compile
    std::cout << str<< std::endl;
    return;
}

template<typename First, typename ... Strings>
void Capitalize_And_Output(First&& str, Strings&&... rest) {
    std::transform(str.begin(), str.end(), str.begin(), ::toupper); //<- will not compile
    std::cout << str << " ";
    Capitalize_And_Output(std::forward<Strings>(rest)...); // don't forget perfect forwarding.
    return;
}

加上在 main 中必须显式使用 std::string 构造函数的事实

    Capitalize_And_Output(std::string("hello"),planet,earth);  // this will work

编辑：不确定如何处理the standard 中描述的这个问题。

Answer 2

http://channel9.msdn.com/Shows/Going+Deep/Cpp-and-Beyond-2012-Scott-Meyers-Universal-References-in-Cpp11

阅读有关通用参考的信息。这是思考这些事情的一种更好的方式（恕我直言），我觉得他的视频很好地解释了右值引用。