On 2008-04-07 13:00:57 -0400, "Bo Persson" <bop@gmb.dk> said:

> Fokko Beekhof wrote:
>> Hello,
>>
>> The following code compiles (gcc/linux/gnu platform)
>> //////////////////////
>> #include <tr1/array>
>>
>> int main()
>> {
>> std::tr1::array<int, 7> a;
>> std::tr1::array<int, 6>::iterator i = a.begin()+1;
>>
>> return 0;
>> }
>> /////////////////////
>>
>> That it compiles is "logical" if you consider that what remains
>> after the first element of a an array of length 7, is an array of
>> length 6.
>>
>> It is also "illogical" if you consider that the iterator returned by
>> a.begin() is of type std::tr1::array<int, 7>::iterator, and merely
>> adding 1 to it should cause it to change its type.
>>
>> Unless, of course, all iterators of type
>> std::tr1::array<int, X>::iterator are the same for all X.
>>
>> Question: is it indeed guaranteed in the standard that all types
>> std::tr1::array<int, X>::iterator are identical for all X ?
>>
>
> No, there are no such guarantees.
>
> That it happens to work is probably an effect of the requirements for
> iterator::value_type, which of course must be int for all
> array<int,X>.
>
>

And, in particular, it's probably because that implementation uses T*
as its iterator type. If it used a nested class (to do bounds checking,
for example), then array<int,6>::iterator and array<int,7>::iterator
would be two different types.

--
Pete
Roundhouse Consulting, Ltd. (www.versatilecoding.com) Author of "The
Standard C++ Library Extensions: a Tutorial and Reference
(www.petebecker.com/tr1book)