Redefining lambdas not allowed in C++11, why?

Question

Example:

#include <functional>  int main() {   auto test = []{};   test = []{};        return 0; } 

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()>&’ 

From the standard 5.1.2.3 (emphasis mine):

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).

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.

Answer 1

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; } 

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