Quick question: Getting an iterator to the last element

[Juha Nieminen]

I'm actually not sure about this one: Does the standard guarantee
that if there's at least one element in the data container, then
"--container.end() " will work and give an iterator to the last
element in the container?

Or is there a cleaner way of getting an iterator to the last
element?

[Claudio A. Andreoni]

Hi!

No, the STL does not guarantees that.
If you call operator-- on an empty container, you get some kind
of junk address (which will probably cause a SIGSEGV).
It is completely upon you to check for the existence of elements
in the container... but if there is any, operator-- will do its
job well.

Remember, if you have an empty data container, the last iterator
is the same as the first (on the GNU STL you have both them set
to 0, if I remember well).
You might so want to check for this condition: x.begin() != x.end()
or x.size() 0 ...

[If you do prefer for some reason accessing the last or the first
element directly (without passing through an iterator) you might
also use front() and back() on some containers (eg: vector).
They return trash if the container is empty anyway].

Bye!
Claudio A. Andreoni
Juha Nieminen wrote:

I'm actually not sure about this one: Does the standard guarantee
that if there's at least one element in the data container, then
"--container.end() " will work and give an iterator to the last
element in the container?

Or is there a cleaner way of getting an iterator to the last
element?

[Ian Collins]

Juha Nieminen wrote:

I'm actually not sure about this one: Does the standard guarantee
that if there's at least one element in the data container, then
"--container.end() " will work and give an iterator to the last
element in the container?

Or is there a cleaner way of getting an iterator to the last
element?

rbegin(), assuming the container supports reverse iterators.

--
Ian Collins.

[Juha Nieminen]

Claudio A. Andreoni wrote:

If you call operator-- on an empty container

Notice that I said this:

Juha Nieminen wrote:

>if there's at least one element in the data container

[Juha Nieminen]

Ian Collins wrote:

rbegin(), assuming the container supports reverse iterators.

But if I need an iterator, not a reverse iterator?

std::list<int>: :iterator iter = l.rbegin();

error: conversion from
`std::reverse_i terator<std::_L ist_iterator<in t, int&, int*'
to non-scalar type
`std::_List_ite rator<int, int&, int*>' requested

[=?ISO-8859-1?Q?Erik_Wikstr=F6m?=]

On 2007-06-17 13:19, Juha Nieminen wrote:

Ian Collins wrote:

>rbegin(), assuming the container supports reverse iterators.

But if I need an iterator, not a reverse iterator?

std::list<int>: :iterator iter = l.rbegin();

error: conversion from
`std::reverse_i terator<std::_L ist_iterator<in t, int&, int*'
to non-scalar type
`std::_List_ite rator<int, int&, int*>' requested

You can use the base() method to get a normal iterator from a reverse
iterator, notice however that if you can base() on l.rbegin() you get
r.end() so you either have to increment the reverse iterator before
calling base(), or decrement the iterator returned by base() to get an
iterator to the last element.

All of this is quite a lot of extra work however, since --l.end() will
give you an iterator to the last element as long as l.size != 0.

Another thing to notice is that all of this (both using --l.end() and
the reverse-iterator) requires that the container supports bidirectional
iterators (though I'm not aware of any containers that don't).

--
Erik Wikström

[Jerry Coffin]

In article <46************ **********@news .song.fi>,
no****@thanks.i nvalid says...

I'm actually not sure about this one: Does the standard guarantee
that if there's at least one element in the data container, then
"--container.end() " will work and give an iterator to the last
element in the container?

Or is there a cleaner way of getting an iterator to the last
element?

container.rbegi n() would be one obvious possibility.

--
Later,
Jerry.

The universe is a figment of its own imagination.

[James Kanze]

On Jun 17, 11:01 am, Juha Nieminen <nos...@thanks. invalidwrote:

I'm actually not sure about this one: Does the standard guarantee
that if there's at least one element in the data container, then
"--container.end() " will work and give an iterator to the last
element in the container?

No. Depending on how the iterator is implemented, it may or may
not be legal. (Obviously, it's never legal if the iterator
isn't at least bi-directional.)

Or is there a cleaner way of getting an iterator to the last
element?

The only sure way is:

C::iterator it = container.end() ;
-- it ;

For containers with random access iterators, container.end() - 1
also works.

--
James Kanze (Gabi Software) email: ja*********@gma il.com
Conseils en informatique orientée objet/
Beratung in objektorientier ter Datenverarbeitu ng
9 place Sémard, 78210 St.-Cyr-l'École, France, [phone]

[Juha Nieminen]

James Kanze wrote:

The only sure way is:

C::iterator it = container.end() ;
-- it ;

Isn't that exactly what I suggested in my question?

[Gavin Deane]

On 18 Jun, 11:08, Juha Nieminen <nos...@thanks. invalidwrote:

James Kanze wrote:

The only sure way is:

C::iterator it = container.end() ;
-- it ;

Isn't that exactly what I suggested in my question?

No. You suggested --container.end() which modifies a temporary - which
is not always legal. The usual example I've seen where your code won't
work is if the iterator is simply a raw pointer.

Gavin Deane