Re: Deleting items from std::list
"Markus Schoder" <a3vr6dsg-usenet@yahoo.de> wrote in message
news:44a19d4b$0$29149$9b4e6d93@newsread4.arcor-online.net...
P.J. Plauger wrote:
"Markus Schoder" <a3vr6dsg-usenet@yahoo.de> wrote in message
news:44a02da0$0$29123$9b4e6d93@newsread4.arcor-online.net...
Howard Hinnant wrote:
In article <1151276851.123572.175280@m73g2000cwd.googlegroups.com>,
"Markus Svilans" <msvilans@gmail.com> wrote:
std::list<int> data;
// [ code here to load data with values ]
std::list<int>::reverse_iterator ri = data.rbegin();
while (ri != data.rend())
{
// Get next iterator in case ri is erased.
std::list<int>::reverse_iterator next = ri;
++next;
// Check if ri needs to be erased
if (*ri == VALUE_TO_ERASE)
{
// [ code here to process *ri before erasing ]
data.erase(ri); // <-- Causes compiler error
}
ri = next;
}
Obviously an erase method is not defined for reverse iterators in
std::list.
The solution below is better but just for completeness sake
data.erase(ri.base());
should work.
for (std::list<int>::iterator ri = data.end(); ri != data.begin();)
{
if (*--ri == VALUE_TO_ERASE)
{
// [ code here to process *ri before erasing ]
ri = data.erase(ri);
}
}
For completeness sake you have to specify how to make it past the
erased element with your reverse iterator. It's way messier than
for Hinnant's direct solution.
Maybe I wasn't clear enough but I meant to say his (Hinnant's)
solution is better. Just wanted to point out that you can erase
through a reverse iterator. The way I did it is however not correct
since &*(ri.base()-1) == &*ri.
Maybe I wasn't clear enough:
1) You're right that erasing the element designated by base() is
off by one. But even if it were right, you'd then have an
invalid base(), useless for finding the next element in the list.
2) Your expression above is ill formed, since base() returns a
bidirectional iterator, which you can't subtract one from.
3) Thus, you're reduced to making a copy of base(), decrementing it,
then using that to erase the designated element.
So my point was that sketching a partial solution, which is
demonstrably hard to get right, doesn't add much to "completeness."
At least we're in violent agreement that Hinnant did it best.
P.J. Plauger
Dinkumware, Ltd.
http://www.dinkumware.com