Re: Bidirectional iterators: signed distance?
Joe Greer wrote:
nottheartistinquestion@hotmail.com wrote in
news:1189010726.097176.232440@r34g2000hsd.googlegroups.com:
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?
I would break this up into two loops and do some more checking than
you do.
I would start with:
int d = 0;
for (Iterator it = i; it != j && it != end(); ++it, ++d);
if (it == j)
return d;
If you incorporate the comparison and return into the loop body,
it would be just a tad faster, I believe:
int d = 0;
for (Iterator it = i; it != end(); ++it, ++d) {
if (it == j)
return d;
}
for (Iterator it = i; it != begin() && it !=j; --it, --d);
if (it !=j)
throw iterators_not_from_same_list();
return d;
Actually the second loop is better in terms of 'i' and 'j' reversed:
d = 0; // did you forget this?
for (Iterator it = j; it != end(); ++it, --d) {
if (it == i)
return d;
}
throw iterators_not ...
I've not tried to run this code, but it should mostly do what you want
and I think it's reasonably clear. Of course, Victor's solution is
cleaner, but will always pay the cost of counting to each of the
iterators. How it is actually used will determine which mechanism is
better.
joe
V
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