Re: Bidirectional iterators: signed distance?
Victor Bazarov wrote:
nottheartistinquestion@hotmail.com wrote:
As an intellectual exercise, I've implemented an STL-esque List<> and
List<>::Iterator. Now, I would like a signed distance between two
iterators which corresponds to their relative position in the list.
For instance, if I did something like distance(list.end(),
list.begin()), I would get -list.size(). The STL's iterator distance
function amounts to something like this:
distance_type distance(Iterator first, Iterator last)
{
distance_type n = 0;
while (first != last) {
++first; ++n;
}
return n;
}
So, if my list has ten elements, and i4 is an iterator at the 4th node
and i2 is an iterator at the 2nd node, then std::distance(i4, i2) is
9. That's meaningless to me. What I want is distance(i4, i2) == -2.
So I implemented distance as a method of List<> like so:
int distance(Iterator i, Iterator j)
{
Iterator tmp = i;
int d = 0;
while (tmp != j) {
if (tmp == end()) {
// oops, try the other way
tmp = i; d = 0;
while (tmp != j) {
--tmp; --d;
}
return d;
}
++tmp; ++d;
}
return d;
}
But this is butt-ugly and I don't like the constant checking for
tmp==end(). Can anyone think of a nicer way to implement this
functionality? Has it been done somewhere else that I can have a look
at?
Find the distance from each of the iterators to the beginning and
then subtract. I.e.
distance(a, b) == distance(begin(), b) - distance(begin(), a)
V
Nothing wrong with this solution but of course the worst case
performance is O(n) where n is the length of the list. You can achieve
O(d), where d is the distance between the two iterators, if you
alternately advance the two iterators until one reaches the original
position of the other. Naturally this won't be as compact as Victor's
solution.
Mark
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