C++ 可变参数模板参数：对象不作为引用转发

Question

假设我有以下代码：

#include <iostream>

struct NotCopyable
{
    std::string value;

    NotCopyable(const std::string& str) :
        value(str)
    {
    }
    
    // This struct is not allowed to be copied.
    NotCopyable(const NotCopyable&) = delete;
    NotCopyable& operator =(const NotCopyable&) = delete;
};

struct Printable
{
    NotCopyable& nc;

    // This struct is constructed with a reference to a NotCopyable.
    Printable(NotCopyable& inNc) :
        nc(inNc)
    {
    }

    // So that std::cout can print us:
    operator const char*() const
    {
        return nc.value.c_str();
    }
};

// This function constructs an object of type T,
// given some arguments, and prints the object.
template<typename T, typename... ARGS>
void ConstructAndPrint(ARGS... args)
{
    T object(std::forward<ARGS>(args)...);
    std::cout << "Object says: " << object << std::endl;
}

int main(int, char**)
{
    // Create a NotCopyable.
    NotCopyable nc("123");

    // Try and construct a Printable to print it.
    ConstructAndPrint<Printable>(nc);    // "Call to deleted constructor of NotCopyable"
    
    // OK then, let's make absolutely sure that we're forwarding a reference.
    NotCopyable& ncRef = nc;

    // Try again.
    ConstructAndPrint<Printable>(ncRef); // "Call to deleted constructor of NotCopyable"
}

似乎 NotCopyable 没有作为参考正确转发，即使我明确提供参考变量作为参数也是如此。为什么是这样？有什么办法可以确保转发的是引用，而不是复制构造的对象？

Answer 1

template<typename T, typename... ARGS>
void ConstructAndPrint(ARGS... args);

按值获取参数，复制也是如此。

您想要转发参考：

// This function constructs an object of type T,
// given some arguments, and prints the object.
template<typename T, typename... ARGS>
void ConstructAndPrint(ARGS&&... args)
{
    T object(std::forward<ARGS>(args)...);
    std::cout << "Object says: " << object << std::endl;
}

C++ 可变参数模板参数：对象不作为引用转发

C++ variadic template arguments: object is not forwarded as a reference

c++

variadic-templates

perfect-forwarding