重新定义lambdas不允许在C ++ 11，为什么？ [英] Redefining lambdas not allowed in C++11, why?

问题描述

示例：

#include <functional>

int main() {
  auto test = []{};
  test = []{};

  return 0;
}

这将在gcc 4.7.2中发出以下错误消息：

This emits the following error message in gcc 4.7.2:

test.cpp: In function ‘int main()’:
test.cpp:5:13: error: no match for ‘operator=’ in ‘test = <lambda closure object>main()::<lambda()>{}’
test.cpp:5:13: note: candidate is:
test.cpp:4:16: note: main()::<lambda()>& main()::<lambda()>::operator=(const main()::<lambda()>&) <deleted>
test.cpp:4:16: note:   no known conversion for argument 1 from ‘main()::<lambda()>’ to ‘const main()::<lambda()>&’

从标准5.1.2.3（强调我）：

From the standard 5.1.2.3 (emphasis mine):

一个实现可以不同于下面描述的定义闭包类型，只要这不改变程序的可观察行为，除非改变：

An implementation may define the closure type differently from what is described below provided this does not alter the observable behavior of the program other than by changing:

- 闭包类型的大小和/或对齐方式，

— the size and/or alignment of the closure type,

- 闭包类型是否可轻易复制（第9条）

— whether the closure type is trivially copyable (Clause 9)

- 闭包类型是否是标准布局类（第9条），

— whether the closure type is a standard-layout class (Clause 9), or

- 闭包类型是否为POD类（第9条）。

— whether the closure type is a POD class (Clause 9).

据我所知，反对。它试图使用已删除的分配运算符和失败。我很想知道是否有一个简单的解决方法，更广泛地说，允许复制可构造性通常省略lambdas的动机理由。

As far as I can tell, this is what I'm running up against. It's attempting to use a deleted assignment operator and failing. I am curious to know if there's an easy workaround, and more broadly what the motivating rationale for allowing copy constructibility to be omitted for lambdas generally.

推荐答案

你似乎认为这两个lambdas有相同的类型，但这不是真的。每个都创建自己的类型：

You seem to think that those two lambdas have the same type, but that is not true. Each one creates its own type:

#include <functional>
#include <type_traits>
#include <iostream>

int main() {
  auto test = []{};
  auto test2 = []{};
  std::cout << std::is_same< decltype( test ), decltype( test2 ) >::value << std::endl;
  return 0;
}

将打印 0 。当然，你从编译器得到的错误消息可能在这方面有点清楚...

will print 0. Of course the error message you are getting from the compiler could be a little bit clearer in this regards...

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