防止复制构造和分配返回值参考 [英] Preventing copy construction and assignment of a return value reference

问题描述

如果我有一个函数返回对我无法控制其源代码的类的实例的引用，请说list<int>:

If I have a function that returns a reference to an instance of a class that I don't have control over its source, say list<int>:

list<int>& f();

我想确保将其值仅分配给另一个引用，例如:

I want to ensure that its value is only assigned to another reference, e.g.:

list<int> &a_list = f();

如果用户要这样做:

list<int> a_list = f(); // note: no '&', so the list is copied

我希望它是一个编译时错误，因为用户将只操作列表的副本而不是原始列表(这绝不是我的应用程序想要/不需要的).

I want it to be a compile-time error since the user would be manipulating only a copy of the list and not the original list (which is never what is intended/wanted for my application).

在上面，有什么方法可以防止复制构造和赋值(例如通过某种包装"类)?

Is there any way to prevent copy-construction and assignment in the above (say via some kind of "wrapper" class)?

理想情况下，如果要使用某些包装类，例如wrapper<T>，我希望它适用于任何类型T的对象.

Ideally, if some wrapper class were to be used, say wrapper<T>, I'd like it to work for objects of any type T.

是的，我知道对于我可以控制的类，我可以简单地将复制构造函数和赋值运算符private设置为:

Yes, I know that for a class that I do have control over, I can simply make the copy-constructor and assignment operator private like:

class MyClass {
public:
    // ...
private:
    MyClass( MyClass const& );
    MyClass operator=( MyClass const& );
};

禁止复制和转让；但是，如上所示，我想对std::list这样做，因为我不能简单地使复制构造函数和赋值运算符private.

to forbid copy-construction and assignment; but, as shown above, I want to do this for, say, std::list for which I can not simply make the copy-constructor and assignment operator private.

推荐答案

这是与我之前的答案不同的答案，因为该问题已得到澄清.我先前的答案可能对有理智要求的人有用，所以我将其保留完整.

This is a separate answer from my previous one, since the problem has been clarified. My previous answer might be useful to someone with sane requirements, so I leave it intact.

所需的行为是不可能的:请求是为了能够一般返回类似于T&的行为，而不是类似于T的行为.必须以某种方式使用户(和编译器！)知道该返回的东西实际上不是引用这一事实.

The desired behavior is impossible: the request is to be able to generically return something that looks exactly like a T& but that does not behave like a T. The fact that this returned thing is NOT actually a reference must be made known the the user (and compiler!) in some way.

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