David Abrahams wrote:
Thorsten Ottosen <tottosen@dezide.com> writes:
David Abrahams wrote:
Thorsten Ottosen <tottosen@dezide.com> writes:
David Abrahams wrote:
Thorsten Ottosen <tottosen@dezide.com> writes:
In that case, this should *not* be integrated with the algorithms, but should instead be provided (if it's needed) as a separate layer that can adjust the endpoints of a range:
drop_front(r) -> new view of r with the begin iterator incremented grow_front(r) -> new view of r with the begin iterator decremented drop_back(r) -> new view of r with the end iterator decremented grow_back(r) -> new view of r with the end iterator incremented
Okay, I see from your points below that the "grow" variants can be dangerous.
Well, all versions can be dangerous: just remember that r can be empty.
No, the "drop" versions are always safe because you can avoid shrinking past empty.
right, my mistake. Now consider having the range [found,end) being returned. How do you supply the functionality given by the return_begin_found flag?
More importantly, to make them safe, you need to supply the original range as a second input. This implies that you need to store a temporary --- this what we are actually trying to avoid.
No, that only applies to the "grow" versions.
right.
If this is an important case (still not convinced),
generally it is hard to say what is important to all people. I do think, however, that splitting a range into different sub-ranges is a fairly common activity. Some ranges are more common:
[begin,found) [found,end)
For that drop_front works great.
how do you construct the range [begin,found) if the algorithm returns ther range [found,end) ?
for example, take unique:
erase( rng, unique(rng) ); // unique should return[found,end) copy( unique<return_begin_found>(rng), out );
but I'm sure that sometimes you want to include the found value in the
I would consider generalizing find with this overload:
find( -1, v, 5 )
That is, "find the element before the one that's equal to 5". And I would *certainly* make it work for forward ranges with somewhat reduced efficiency.
I'm not that keen on providing to many new algoriths.
What do you think this thing called "find" with the explicit template parameter is, if not a new algorithm?!
It's a utility wrapper of an existing std algorihtm.
The return-value mechanism is a more like a layer on top of the algorithm, not a new algorithm.
How is that distinct from the much simpler find illustrated above? Are you saying that the use of the explicit template argument is what makes it "not a new algorithm?"
You'l need to implement a new version of find so that it works for forward iterators, you don't just forward to an existing.
I don't think it's necessary to have the policy be a compile-time thing. But if it were, I'd still pass it by by value.
it is necessary if you want the algorithm to be able to return either an iterator or a range.
I don't want that, but please explain.
For eaxmple, take unique(): template< range_return_value re, class Rng > inline BOOST_DEDUCED_TYPENAME range_return<Rng,re>::type unique( Rng& r ) { return range_return<Rng,re>::pack( std::unique(boost::begin(r), boost::end(r)), r ); } this allows the algorithm to return an iterator like the old version. It also illustrates that if we change the template parameter to a function parameter, pack() becomes a big function with a switch case.
if only ranges are supported, using a parameter seems like the thing
^ regular run-time function---------+ ??
right.
Anyway, how do you check for success? What if 5 is the first element? Do you get an empty range or do you just get v back?
you get an empty range back if nothing is found:
Of course you do. That wasn't my question. In the original example you were searching for 5. So what happens when you ask for the range beginning just before (or just after, for that matter) 5 in a single-element range containing just 5?
then you get an empty range. [Dave:]
Anyway, how do you check for success? What if 5 is the first element? Do you get an empty range or do you just get v back?
if( find( v, 6 ) ) { ... }
Thus we would be forced to create a temporary.
yes, but only if you're not passing the range directly to another algorithm.
Now you're arguing with yourself. I'm happy to sit back and see who wins this one.
You asked how you check for success. My reply was the if-statement above. If you ask for a range [begin,found) when you search for 5, and 5 is the fist element, then of course you get back an empty range. -Thorsten