I am trying to write an adaptor function that takes a string and a
list of character delimiters and populates an output iterator with the
result. Specifically, this is what I have:
using boost::algorithm::make_split_iterator;
using boost::algorithm::split_iterator;
using boost::iterator_range;
using boost::copy_range;
using boost::algorithm::is_any_of;
using boost::algorithm::token_finder;
template
inline OutIt split_and_copy(const std::basic_string &source,
const D &delimeters,
OutIt out)
{
typedef std::basic_string mystring;
typedef split_iteratormystring::const_iterator SplitIt;
typedef iterator_rangemystring::const_iterator RangeIt;
const SplitIt b = make_split_iterator(source,
token_finder(is_any_of(delimeters)));
const SplitIt e;
return std::transform(b, e, out, copy_range);
}
This compiles but strikes me as a bit verbose. I mean, am I missing
something visa-vi the proper interaction of split_iterator,
iterator_range and the container they are splitting and making a range
of?
I believe since b is a split_iterator type, that *b is of type
iterator_range, is this not correct? The templates would be as I
listed for SplitIt and RangeIt, as I understand it. Now, originally I
was hoping there would be an automatic conversion back from
iterator_range to the ranged container, which would make a copy of the
elements in the range and stick them into a container of the original
type (in this case, string). Is there no conversion from
iterator_rangemystring::const_iterator back to mystring? You see,
originally, I wanted to write:
std::copy(b, e, out);
But it seems there is no way around using std::transform with an
explicit conversion from RangeIt to mystring with copy_range. Or am I
wrong in that assumption? No, I don't want to use the split function
since I don't want to populate the whole container, just write my
entries to an output iterator; it's generic in that sense to allow for
ostream_iterator or back_inserters or whatever. Also, is there some
way to not have to explicitly provide the copy_range input type since
that type is already known to be the same type as *b and *e?
Thanks for any insight you might be able to offer!
Jeffrey
--
"Scan not a friend with a microscopic glass; you know his faults so
let his foibles pass." -- Victorian Proverb, likely Sir Frank Crisp,
1st Baronet of Bungay
~,-;` The TimeHorse
