Re: Bidirectional iterators: signed distance?
Mark P wrote:
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.
I am by no means the master of the Big-O notation, but it seems that
O(n) and O(d) are the same terms. The worst case scenario with my
original suggestion is when 'a'=='end()' && 'b'=='end()'. Then the
number of comparisons is 2*N. The worst case scenario with counting
in both directions is when 'a'=='end()' && 'b'=='begin()', i.e. the
distance is -N, in which case the number of comparisons is N, still.
So, you get about 50% savings on iterator comparisons with going in
two directions.
V
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