在lambda中使用未捕获的变量 [英] Using of not captured variable in lambda

问题描述

我不太理解C ++ 14标准草案 N4140 5.1.2.12 [expr.prim.lambda] 中的示例。

I can not quite understand an example from C++14 standard draft N4140 5.1.2.12 [expr.prim.lambda].

具有关联的capture-default的lambda表达式，它不会显式地捕获此值或具有自动存储持续时间的变量（这不包括已发现要引用的任何id表达式）到initcapture的关联非静态数据成员），如果复合语句被称为隐式捕获实体（即，此变量或变量）：

A lambda-expression with an associated capture-default that does not explicitly capture this or a variable with automatic storage duration (this excludes any id-expression that has been found to refer to an initcapture’s associated non-static data member), is said to implicitly capture the entity (i.e., this or a variable) if the compound-statement:

• odr-使用实体，或


• 在可能评估的表达式中命名实体，其中封闭的全表达式取决于在范围内声明的通用lambda参数。 lambda表达式。

[示例：

void f(int, const int (&)[2] = {}) { } // #1
void f(const int&, const int (&)[1]) { } // #2
void test() {
  const int x = 17;
  auto g = [](auto a) {
    f(x); // OK: calls #1, does not capture x
  };
  auto g2 = [=](auto a) {
    int selector[sizeof(a) == 1 ? 1 : 2]{};
    f(x, selector); // OK: is a dependent expression, so captures x
  };
}

-结束示例]

所有此类隐式捕获的实体都应在lambda表达式的作用域内声明。

All such implicitly captured entities shall be declared within the reaching scope of the lambda expression.

[注意：嵌套的lambda-expression隐式捕获实体可能会通过包含的lambda-expression导致其隐式捕获（请参见下文）。隐式使用odr会导致隐式捕获。 —尾注]

[ Note: The implicit capture of an entity by a nested lambda-expression can cause its implicit capture by the containing lambda-expression (see below). Implicit odr-uses of this can result in implicit capture. —end note ]

我认为短语带有相关捕获的lambda表达式的开头- default 应该禁止任何隐式捕获（并通过评论确认），因此＃1 调用将导致错误（有关使用未捕获的变量的某些信息） ）。那么它是如何工作的呢？ f 的第一个参数是什么？如果退出 test（）范围后将调用 g 怎么办？如果我将＃1 签名更改为 void（const int&）怎么办？

I thought that the beginning of a phrase a lambda-expression with an associated capture-default should prohibit any implicit capture (and it's confirmed by comment), therefore #1 call will lead to an error (something about using not captured variable). So how it works? What will be first argument of f? What if g will be called after exiting test() scope? What if I change #1 signature to void(const int&)?

-

upd：感谢大家对它的工作原理的解释。稍后，我将尝试查找并发布有关此案例的标准参考。

upd: Thanks to all for explanation of how it works. Later I'll try to find and post references to standard about this case.

推荐答案

作为T.C。在他的评论中说，＃1不需要捕获，因为 x 在编译时是已知的，因此被烘焙到lambda中。与在编译时如何知道函数 f 一样，因此不需要捕获它。

As T.C. said in his comment, #1 does not require a capture as x is known at compile time and is therefore baked into the lambda. Not unlike how the function f is known at compile time so it doesn't need to be captured.

I相信如果将 f 的签名更改为 int const& ，您现在正在尝试传递常量的地址
。因此，它可能会发生变化，因此需要按值或引用捕获 x 。

I believe if you change f's signature to int const & you are now attempting to pass the address of the constant which is on the stack, thus subject to changes, and it would require capturing x by value or reference.

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