Re: [boost] [range] RFC: span(iterator, size)

Subject: Re: [boost] [range] RFC: span(iterator, size)
From: TONGARI J (tongari95_at_[hidden])
Date: 2013-10-30 03:29:23

2013/10/30 TONGARI J <tongari95_at_[hidden]>

> 2013/10/30 Evgeny Panasyuk <evgeny.panasyuk_at_[hidden]>
>
>>  1. So, span(first, n) and span(next(first), n) will result in "equal"
>> iterators?
>>
>
> No.
>

if you mean begin-or-end(span(first, n)) == begin-or-end(span(next(first),
n)), then yes, but it's pathic.

> Suppose we have a sequence S = {1, 2, 3, 4, 5}, &i is the iterator to i.
> Given first = &1, n = 4,
> span(first, n)->[begin {.it = &1, .count = 0}, end { .it = &1, .count = 4
> }], which results in {1, 2, 3, 4}
> span(next(first), n)->[begin {.it = &2, .count = 0}, end { .it = &2,
> .count = 4 }], which results in {2, 3, 4, 5}
>
>
>> 2. As I understand, span(first, n) will produce following "iterators":
>> begin {.it = first, .count = 0}
>> end { .it = ???, .count = n }
>
> What is end.it?
>
>
> end { .it = first, .count = n }
>
>
>> Especially you mentioned that you plan to preserve same range category as
>> original iterator has - what about BidirectionalIterators?
>
>
> That's my mistake, they should be ForwardIterators.
>
>