Iterator to last element in list?

[Howard]

Is there an easy way to get an iterator (*not* a reverse-iterator) to the
last element in a list? The last() function returns the element itself, not
an iterator.

Thanks,
-Howard

 

[Ian Collins]

Is there an easy way to get an iterator (*not* a reverse-iterator) to
the last element in a list? The last() function returns the element
itself, not an iterator.

How about list.end()-1 or list.start()+(list.size()-1)

 

[Mark P]

How about list.end()-1 or list.start()+(list.size()-1)

list iterators are bidirectional, not random access, so you can't do
arithmetic like this. Not to mention that list.size() is O(n).

But you can do:

list<T>::iterator it = l.end();
if( !l.empty())
--it;

Now "it" points to either the last element or, if the list is empty, end().

 

[Ian Collins]

list iterators are bidirectional, not random access, so you can't do
arithmetic like this. Not to mention that list.size() is O(n).

Oops, sorry.

 

[Howard]

list iterators are bidirectional, not random access, so you can't do
arithmetic like this. Not to mention that list.size() is O(n).

But you can do:

list<T>::iterator it = l.end();
if( !l.empty())
--it;

Now "it" points to either the last element or, if the list is empty,
end().

Yeah, that's what I'm doing now, but I was hoping for a single expression I
could use. This is working fine.

Thanks,
-Howard

 

[Daniel T.]

Yeah, that's what I'm doing now, but I was hoping for a single expression I
could use. This is working fine.

assert( !l.empty() );
list<T>::iterator it = --l.end();

 

[Daniel T.]

Yeah, that's what I'm doing now, but I was hoping for a single expression I
could use. This is working fine.

Or how about:

assert( !l.empty() );
list<T>::iterator it = l.rend().base();

 

[Kai-Uwe Bux]

assert( !l.empty() );
list<T>::iterator it = --l.end();

I prefer your other idea:

list<T>::iterator it = l.rbegin().base();

This is more generic, e.g. you can use it in algorithms templated on
containers:

template < typename Container >
typename Container::iterator
last_iter ( Container & c ) {
return ( c.rbegin().base() );
}

will always work whereas

template < typename Container >
typename Container::iterator
last_iter ( Container & c ) {
return ( -- c.end() );
}

can fail for std::vector where the iterator type might be a naked pointer.



Best

Kai-Uwe Bux

 

[Daniel T.]

Isn't rend().base() the beginning of the list?

Right, sorry. rbegin().base().

 

[Pete Becker]

assert( !l.empty() );
list<T>::iterator it = --l.end();

Maybe, but it's not required to work. l.end() returns a temporary, and
-- isn't required to be a member function. It usually is, so this will
work with most implementations.

list<T>::iterator it = l.end();
--it;

 

[Ben Rudiak-Gould]

list<T>::iterator it = l.rbegin().base();

This won't work; rbegin().base() is equivalent to end().

-- Ben

 

[Kai-Uwe Bux]

This won't work; rbegin().base() is equivalent to end().

Ah, right.


Thanks

Kai-Uwe Bux

 

[Mark P]

Or how about:

assert( !l.empty() );
list<T>::iterator it = l.rend().base();

Isn't rend().base() the beginning of the list?