为什么默认情况下，lambda函数会丢弃推导的返回类型引用? [英] Why do lambda functions drop deduced return type reference by default?

问题描述

在C ++ 14中，为什么默认情况下具有推导的返回类型的lambda函数会从返回类型中删除引用? IIUC，因为具有推导的返回类型(没有显式的尾随返回类型)的C ++ 14 lambda函数的返回类型为auto，这会删除引用(以及其他内容).

In C++14, why do lambda functions with a deduced return type drop references from the return type by default? IIUC, since C++14 lambda functions with a deduced return type (without an explicit trailing return type) have a return type of auto, which drops references (among other things).

为什么要做出这个决定?在我看来，当那是您的return语句返回的内容时，删除引用是一个陷阱.

Why was this decision made? It seems to me like a gotcha to remove a reference when that's what your return statement returns.

此行为对我造成了以下讨厌的错误:

This behavior caused the following nasty bug for me:

class Int {
public:
   Int(int i) : m_int{i} {}
   int m_int;
};

class C {
public:
    C(Int obj) : m_obj{obj} {}
    const auto& getObj() { return m_obj; }
    Int m_obj;
};

class D {
public:
    D(std::function<const Int&()> f) : m_f{f} {}
    std::function<const Int&()> m_f;
};

Int myint{5};
C c{myint};
D d{ [&c](){ return c.getObj(); } } // The deduced return type of the lambda is Int (with no reference)
const Int& myref = d.m_f(); // Instead of referencing myint, myref is a dangling reference; d.m_f() returned a copy of myint, which is subsequently destroyed.

在初始化d时指定所需的返回类型可解决此问题:

Specifying the desired return type when initializing d resolves the issue:

D d{ [&c]() -> const Int& { return c.getObj(); } }

有趣的是，即使auto返回类型推论有意义，不是std::function<const Int&>用返回非引用的函数来快乐地初始化std::function<const Int&>的错误吗?我也通过明确编写来看到这一点:

Interestingly, even if the auto return type deduction makes sense, isn't it a bug that std::function<const Int&> gets happily initialized with a function that returns a non-reference? I see this also by writing explicitly:

D d{ [&c]() -> Int { return c.getObj(); } }

编译没有问题. (在Xcode 8，clang 8.0.0上)

which compiles without a problem. (on Xcode 8, clang 8.0.0)

推荐答案

我认为您绊脚的地方实际上是return c.getObj();行中的表达式c.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()具有类型const Int&.但是，这是不正确的.表达式从不具有引用类型.正如Kerrek SB在评论中所指出的那样，有时我们会将表达式视为具有引用类型，这是节省冗长的捷径，但这会导致误解，因此，我认为了解实际情况很重要.

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(包括作为在getObj声明中的返回类型)会影响被声明的事物的初始化方式，但是一旦被初始化，就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.

这是一个更简单的示例:

Here is a simpler example:

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

与

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

这两个代码与^{完全相同(除了decltype(a)或decltype(b)的结果，这对系统来说有点麻烦)}.在这两种情况下，表达式a和b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"并且b是某种伪装的指向a的指针的情况并非如此.他们都处于平等地位.这是一个有两个名称的对象.

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:除了访问权限外，表达式c.getObj()的行为与c.m_obj完全相同.类型为Int，值类别为"lvalue".返回类型为getObj()的&仅指示这是一个左值，并且还将指定一个已经存在的对象(大约来说).

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();推导的返回类型与return c.m_obj;相同，后者与模板类型推导兼容(如其他地方所述)，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.

NB.如果您了解了这篇文章，您还将理解为什么我不喜欢将引用"的教学法讲授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.

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