是否有可能不允许引用对象 [英] Is it possible to disallow taking a reference to an object

问题描述

我想做一个相反的事情，使一个类的实例不可复制，即确保一个特定类的实例只能被传递作为副本，而不作为参考。如果有任何函数试图通过引用接收它，我希望它给出一个编译错误（理想情况下）或运行时错误。

我不认为制作 operator& private会这样做，是否有合法的方法。

解决方案

我不认为让运算符&

否，因为& ; 在函数签名中用于通过引用的不是操作员。您是在谈论地址运算符（一元）或按位与运算符（二进制）。因此，它与传递引用无关。

无法禁止类型的传递引用。 / p>

我怀疑您的动机是否足够强大，并且您似乎对传递机制缺乏理解：

如果任何函数试图通过引用接收它，我希望它给出编译错误（理想情况下）或运行时错误。

一个函数通过引用或值传递参数。它由声明决定，我想您的困惑源于此。例如：

  void foo（X x）; 
  

采用值 x 。无法通过引用传递它。没门。同样：

  void foo（X& x）
  

将其作为参考，并且始终如此。

I want to do the opposite of making instances of a class noncopyable, that is, make sure that instances of a particular class can be passed only as a copy and not as a reference. If any function tries to receive it by reference, I would like it to give a compilation error (ideally) or a run time error.

I dont think making operator & private is going to do it, is there a legitimate way of doing this.

解决方案

I dont think making operator & private is going to do it, is there a legitimate way of doing this.

No, because the & you use in function signatures for pass-by-reference is not an operator. You're talking either about the address-of operator (unary) or bitwise-and operator (binary). So it has nothing to do with pass-by-reference.

There's no way to disallow pass-by-reference for a type.

I doubt your motivation is strong enough to do this, and you appear to have a bad understanding of the passing mechanism:

If any function tries to receive it by reference, I would like it to give a compilation error (ideally) or a run time error.

A function either passes a parameter by reference, or by value. It's decided by its declaration, and I think your confusion stems from here. For example:

void foo(X x);

takes the parameter x by value. There's no way to pass it by reference. No way. Likewise:

void foo(X& x)

takes it by reference, and it always will.

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