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为什么 lambda 函数默认删除推导的返回类型引用?

更新时间:2023-11-05 17:51:46

我认为你绊倒的地方实际上是在 return c 行中使用表达式 c.getObj().getObj();.

I think the place you are stumbling is actually with the expression c.getObj() in the line return c.getObj();.

您认为表达式 c.getObj() 的类型为 const Int&.然而事实并非如此.表达式永远没有引用类型.正如 Kerrek SB 在评论中指出的那样,我们有时将表达式视为具有引用类型,作为节省冗长的捷径,但这会导致误解,因此我认为了解真正发生的事情很重要.

You think the expression c.getObj() has type const Int&. However that is not true; expressions never have reference type. As noted by Kerrek SB in comments, we sometimes talk about expressions as if they had reference type, as a shortcut to save on verbosity, but that leads to misconceptions so I think it is important to understand what is really going on.

在声明中使用引用类型(包括在 getObj 的声明中作为返回类型)会影响被声明的事物的初始化方式,但是一旦初始化,就没有不再有任何证据表明它最初是一个参考.

The use of a reference type in a declaration (including as a return type as in getObj's declaration) affects how the thing being declared is initialized, but once it is initialized, there is no longer any evidence that it was originally a reference.

这是一个更简单的例子:

Here is a simpler example:

int a; int &b = a;  // 1

对比

int b; int &a = b;  // 2

这两个代码完全相同 (除了 decltype(a)decltype(b) 的结果,这对系统).在这两种情况下,表达式 ab 都具有类型 int 和值类别lvalue"并表示相同的对象.a 不是真实对象"而 b 是某种指向 a 的伪装指针.他们是平等的.这是一个有两个名字的对象.

These two codes are exactly identical (except for the result of decltype(a) or decltype(b) which is a bit of a hack to the system). In both cases the expressions a and b both have type int and value category "lvalue" and denote the same object. It's not the case that a is the "real object" and b is some sort of disguised pointer to a. They are both on equal footing. It's one object with two names.

现在回到您的代码:表达式 c.getObj()c.m_obj 具有完全相同的行为,除了访问权限.类型为Int,值类别为lvalue".getObj() 的返回类型中的 & 仅指示这是一个左值,并且它还将指定一个已经存在的对象(大致来说).

Going back to your code now: the expression c.getObj() has exactly the same behaviour as c.m_obj, apart from access rights. The type is Int and the value category is "lvalue". The & in the return type of getObj() only dictates that this is an lvalue and it will also designate an object that already existed (approximately speaking).

因此从 return c.getObj(); 推导出的返回类型与 return c.m_obj; 的相同,这 - 是兼容的使用模板类型推导,如别处所述——不是引用类型.

So the deduced return type from return c.getObj(); is the same as it would be for return c.m_obj; , which -- to be compatible with template type deduction, as mentioned elsewhere -- is not a reference type.

注意.如果你理解了这篇文章,你就会明白为什么我不喜欢将引用"的教学法教为自动取消引用的伪装指针",这介于错误和危险之间.

NB. If you understood this post you will also understand why I don't like the pedagogy of "references" being taught as "disguised pointers that auto dereference", which is somewhere between wrong and dangerous.