为什么 lambda 函数默认删除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，因为具有推导返回类型(没有显式尾随返回类型)的 C++14 lambda 函数的返回类型为 auto，这会删除引用(除其他外).

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).

为什么会做出这个决定?在我看来，当您的 return 语句返回时删除引用就像一个陷阱.

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

这种行为给我带来了以下令人讨厌的错误:

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解决了问题:

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 返回类型推导是有意义的，难道 std::function 被愉快地初始化时难道不是一个错误吗?一个返回非引用的函数?我也通过明确的写作看到了这一点:

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)

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

推荐答案

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) 或 decltype(b) 的结果，这对系统)}.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"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() 与 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.

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