Re: [boost] [iterator] function_output_iterator constructed from a lambda function is not assignable

On Sunday, February 14, 2016 at 6:28:47 PM UTC-6, Mikhail Matrosov wrote:

Consider the following code snippet:

auto f = [](int x) { std::cout << x; };auto it = boost::make_function_output_iterator(f);decltype(it) it2 = it; // Ok, copied it2 = it; // Does not compile, cannot assign!

The problem is, function_output_iterator constructed in this way is not assignable, and thus does not satisfy the Iterator http://en.cppreference.com/w/cpp/concept/Iterator concept, which requires type to be CopyAssignable http://en.cppreference.com/w/cpp/concept/CopyAssignable.

One way to deal with this is to put it into boost::optional. However, it now needs to be dereferenced. This can be easily done with `fit::indirect`: template<class F> fit::indirect_adaptor regular(F f) { return fit::indirect(boost::make_optional(f)); } See http://fit.readthedocs.org/en/latest/indirect/index.html Another option is to use `std::ref`, but this requires carefully managing the lifetime of the lambda so it lives as long as the iterator. In some circumstances this is doable. Paul