Subject: Re: [boost] [iterator] function_output_iterator constructed from a lambda function is not assignable
From: Paul Fultz II (pfultz2_at_[hidden])
Date: 2016-02-14 21:59:20
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<boost::optional<F>> 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