如何获取通过可变参数构造函数调用的复制构造函数？

问题描述

在下面的代码中，可变参数构造函数被调用了两次。如何在适当的时候获取被调用的复制构造函数，而不是可变参数构造函数的单个参数版本？

In the following code, the variadic constructor is called twice. How can I get the copy constructor to be called instead of the single argument version of the variadic constructor when appropriate?

#include <iostream>

struct Foo
{
    Foo(const Foo &)
    {
        std::cout << "copy constructor\n";
    }

    template<typename... Args>
    Foo(Args&&... args)
    {
        std::cout << "variadic constructor\n";
    }

    std::string message;
};

int main()
{
    Foo f1;
    Foo f2(f1); // this calls the variadic constructor, but I want the copy constructor.
}

推荐答案

其中构造函数是可变的。具有非可变构造函数模板的以下类具有相同的行为：

This actually has nothing to do with the fact that the constructor is variadic. The following class with a non-variadic constructor template exhibits the same behavior:

struct Foo
{
    Foo() { }

    Foo(const Foo& x)
    {
        std::cout << "copy constructor\n";
    }

    template <typename T>
    Foo(T&& x)
    {
        std::cout << "template constructor\n";
    }

};

问题是构造函数模板是更好的匹配。要调用复制构造函数，需要进行限定转换，以将非常量值 f1 绑定到 const Foo& 必须添加const限定符）。

The problem is that the constructor template is a better match. To call the copy constructor, a qualification conversion is required to bind the non-const lvalue f1 to const Foo& (the const qualification must be added).

要调用构造函数模板，不需要转换： T 推导出 Foo& ，在引用折叠（ Foo&&& - > Foo& ），提供参数 x type Foo& 。

To call the constructor template, no conversions are required: T can be deduced to Foo&, which after reference collapsing (Foo& && -> Foo&), gives the parameter x type Foo&.

你可以通过提供一个具有非常量引用参数 Foo& 的第二个拷贝构造函数来解决这个问题。

You can work around this by providing a second copy constructor that has a non-const lvalue reference parameter Foo&.

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