c++ - 在 C++11 中不允许重新定义 lambda，为什么？

Question

例子:

#include <functional>

int main() {
  auto test = []{};
  test = []{};
    
  return 0;
}

这会在 gcc 4.7.2 中发出以下错误消息:

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()>&’

来自标准 5.1.2.3(强调我的):

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:

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

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

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

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

据我所知，这就是我遇到的问题。它试图使用已删除的赋值运算符并失败。我很想知道是否有一个简单的解决方法，以及更广泛地说，允许 lambdas 通常省略复制可构造性的动机是什么。

Answer 1

您似乎认为这两个 lambda 具有相同的类型，但事实并非如此。每个都创建自己的类型:

#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信息在这方面可能会更清楚一点......