Re: STL list code fires assertion in VC++ 2005, but not in VC++ 2003.
Jason Doucette wrote:
Sorry, you mean the newsgroup FAQ, obviously, duh. I had forgotten
which newsgroup I was replying to.
...which is parashift.com's C++ FAQ Lite, which is not a FAQ just for
this group.
5.8 is about posting non-working code questions:
http://www.parashift.com/c++-faq-lite/how-to-post.html#faq-5.8
Is this what you mean?
If so, I am not asking anyone to fix my code. I'm looking for a
resource (that I can research myself) about what changes were made to
the STL implementation from VC++ 2003 to VC++ 2005, so I can determine
what I'm now doing wrong.
Thanks,
Jason
Victor was almost certainly trying to hint that you should post some
code. Without seeing code that demonstrates the error, the best we can
do is guess.
From the rest of this thread, it appears to me that you are under the
impression that iterators are pointers, which is not (necessarily) true.
Setting an iterator to NULL, for example, doesn't make any sense. You
should treat an iterator as an opaque object that can only be used in
the expressions guaranteed by the standard. If you don't have a copy of
the standard, the following is reasonably close to correct:
http://www.sgi.com/tech/stl/table_of_contents.html
The equivalent of a NULL pointer for iterators is the past-the-end
iterator (what you get from container.end()). The correct way to do
what you described elsewhere in the thread would look something like this:
template <typename T>
typename std::list<T>::iterator some_algorithm(
const std::list<T> & lst, const T & value)
{
// Do whatever and return an iterator.
// Not found, return the past the end iterator.
return lst.end();
}
void some_function()
{
std::list<int> lst;
// ...
std::list<int>::iterator i = some_algorithm(lst, 42);
if (i == lst.end())
{
// Handle not found case.
}
}
--
Alan Johnson