Re: random access iterator

=?Utf-8?B?R2Vvcmdl?= <>
Mon, 3 Mar 2008 22:47:10 -0800
Thanks Alexander,

Your algorithm is quite smart than traditional bubble sort implementation,
which is different form traditional implementation, like this,

Your algorithm has a status to mark whether in last round there is any
changes, if no change, just quit the algorithm. For the traditional
implementation, it also has advantage which we do not need to iterate from
beginning every time as you did in your solution.

So, my question is, compared with the traditional implementation, in what
situations your algorithm is better/worse? Why?


"Alexander Nickolov" wrote:

Well, it can be useful for example when writing bubble-sort
by hand (not that I'd actually advocate doing that...):

bool bSwap = false;
while (did_swap) {
    bSwap = false;
    for (range_type::iterator it = range.begin(); it != range.end() && it+1
!= range.end(); ++it) {
        if (*it > it[1]) {
            std::swap(*it, it[1]);
            bSwap = true;

The important point is about accessing elements relative to the current
position of the iterator, not the bubble-sort algorithm of course... :)

Alexander Nickolov
Microsoft MVP [VC], MCSD

"Abhishek Padmanabh" <> wrote in message

"George" <> wrote in message

For random access iterator, operator[] is supported. Mentioned in
book, Chapter 19 (Iterators and Allocators).

I have not used operator[] on random access iterator before and I have
found a good and simple sample either. :-)

That is because it is not very intuitive to use as compared to using the
operator[] overloaded for the container itself. Also, as far as VC++ 2005
is concerned, it causes two dereferences. You can use the iterator's
operator[] as below:

using std::vector;
int main()
 vector<int> vec(10);
 vector<int>::iterator it = vec.begin();
 int fifth_element = it[4]; //equivalent to advance it by 4 and
dererefence the resulting iterator
 int fifth_element = vec[4]; //preferred and more intuitive

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