iterator range

[subramanian100in@yahoo.com, India]

Whenever we specify an iterator range to a container operation or an
std::algorithm, the first iterator should always be a valid iterator
ie it should point to some element in the respective container? Am I
correct?

For example, suppose

vector<stringv;

v.erase(v.begin (), v.end());
sort(v.begin(), v.end());

Here since 'v' is empty, v.begin() does not refer to an element in the
container. So
both the above statements invoke undefined behaviour. Am I correct ?

Kindly clarify.

Thanks
V.Subramanian

[Andrey Tarasevich]

Re: iterator range

subramanian100i n@yahoo.com, India wrote:

Whenever we specify an iterator range to a container operation or an
std::algorithm, the first iterator should always be a valid iterator
ie it should point to some element in the respective container? Am I
correct?
...

Yes and no.

Yes, the first iterator has to be valid. Moreover, the second iterator also has
to be valid. Both of them have to be valid.

And no, a valid iterator does not necessarily point to the existing element in
the container. The iterator that points to the imaginary element that follows
the last existing element in the container is also perfectly valid. It cannot be
dereferenced, but it still can be used with non-dereferencing iterator
operations. When defining a sequence, the first iterator (begin) can be equal to
the second iterator (end), which defines an empty sequence. In this case there's
no requirement for the first iterator to point to an existing element.

--
Best regards,
Andrey Tarasevich

[subramanian100in@yahoo.com, India]

Re: iterator range

* On Feb 9, 1:12 am, "Alf P. Steinbach" <al...@start.no wrote:

* On Feb 9, 2:36 am, Andrey Tarasevich <andreytarasev. ..@hotmail.com>
wrote:

Thanks to both Alf P. Steinbach and Andrey Tarasevich for the reply:

(Please bear with me for the following post)

From your replies above, what I understand is that:

when we specify an iterator range, to a container operation or an
std::algorithm, both the first iterator and the last iterator can be
the sentinel ie both of them can be off-the-end iterators. correct?

If so, how do we know whether the implementation of container member
operations or the std:algorithms which accept an input range, will not
dereference the first iterator when both the first and last are
sentinel.

Essentially how do we know that the following operations' definitions
on all compiler implementations will not dereference the the first
iterator which is already the sentinel because the container is empty.

vector<stringv;

v.erase(v.begin (), v.end());
sort(v.begin(), v.end());

Kindly clarify.

Thanks
V.Subramanian

[James Kanze]

Re: iterator range

On Feb 9, 12:34 pm, "subramanian10. ..@yahoo.com, India"
<subramanian10. ..@yahoo.comwro te:

From your replies above, what I understand is that:

when we specify an iterator range, to a container operation or an
std::algorithm, both the first iterator and the last iterator can be
the sentinel ie both of them can be off-the-end iterators. correct?

For all of the algorithms in the standard library, at any rate.

If so, how do we know whether the implementation of container
member operations or the std:algorithms which accept an input
range, will not dereference the first iterator when both the
first and last are sentinel.

The same way that you know that it won't start by incrementing
the begin iterator three times, even if there are only two
elements in the sequence. The specifications of the function.
(Presumably, you could write a function which has a
pre-condition that distance( begin, end ) >= 3. There are
probably even some strange cases where it makes sense.)

The specifications of all of the functions in the standard
library, and most reasonably defined user functions, allows
empty sequences.

Essentially how do we know that the following operations'
definitions on all compiler implementations will not
dereference the the first iterator which is already the
sentinel because the container is empty.

vector<stringv;

v.erase(v.begin (), v.end());
sort(v.begin(), v.end());

Because the specifications of the functions in the standard say
so.

--
James Kanze (GABI Software) email:james.kan ze@gmail.com
Conseils en informatique orientée objet/
Beratung in objektorientier ter Datenverarbeitu ng
9 place Sémard, 78210 St.-Cyr-l'École, France, +33 (0)1 30 23 00 34

[James Kanze]

Re: iterator range

On Feb 9, 2:24 pm, Pete Becker <p...@versatile coding.comwrote :

[...]

Essentially how do we know that the following operations' definitions
on all compiler implementations will not dereference the the first
iterator which is already the sentinel because the container is empty.

vector<stringv;

v.erase(v.begin (), v.end());
sort(v.begin(), v.end());

We know it won't do that because we trust the implementor of sort to
understand the rules for sequences.

We know it doesn't do that because the standard doesn't allow it
to do that. The standard pretty much says (not in that
language) that none of the algorithms will ever access through
an iterator equalt to end.

Of course, there is an aspect of trust involve. We trust the
implementors to respect the specifications of what they
implement.

--
James Kanze (GABI Software) email:james.kan ze@gmail.com
Conseils en informatique orientée objet/
Beratung in objektorientier ter Datenverarbeitu ng
9 place Sémard, 78210 St.-Cyr-l'École, France, +33 (0)1 30 23 00 34

[Pete Becker]

Re: iterator range

On 2008-02-09 14:01:48 -0500, James Kanze <james.kanze@gm ail.comsaid:

On Feb 9, 2:24 pm, Pete Becker <p...@versatile coding.comwrote :
>
[...]

>>Essentially how do we know that the following operations' definitions
>>on all compiler implementations will not dereference the the first
>>iterator which is already the sentinel because the container is empty.

>

>>vector<string v;

>

>>v.erase(v.beg in(), v.end());
>>sort(v.begin( ), v.end());

>

>We know it won't do that because we trust the implementor of sort to
>understand the rules for sequences.

>
We know it doesn't do that because the standard doesn't allow it
to do that. The standard pretty much says (not in that
language) that none of the algorithms will ever access through
an iterator equalt to end.

Yes, that's one of the rules for sequences.

>
Of course, there is an aspect of trust involve. We trust the
implementors to respect the specifications of what they
implement.

Yes, that's another way of saying what I said.

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