From: Daniel Spangenberg (dsp_at_[hidden])
Date: 2003-10-31 05:30:43
Rob Stewart schrieb:
> From: "Peter Dimov" <pdimov_at_[hidden]>
> > Have you looked at the signature of std::nth_element? ;-)
Yes I have, What is the problem?
> Actually, I searched to find it and then skimmed the description,
> assuming that Daniel had looked more closely. I didn't look at
> the signature until you questioned me.
>
> As it turns out, std::nth_element() operates strictly with
> iterators and returns void! Doh!
>
> Given my now enlightened understanding of nth_element, I don't
> consider it analogous to find_nth() at all, so you can consider
> all of my capitulation above rescinded!
>
> Choosing a different name will resolve the problem as previously
> stated.
Maybe a misunderstanding on my side. I did not want to say, that
the complete signature of nth_element is related to the find_nth
signature nor that we should adopt that signature. I only wanted to
show that nth_element actually means (n - 1)th element for normal
speakers. Given the example from Austern's book
int array[] = {7, 2, 6, 11, 9, 3, 12, 10, 8, 4, 1, 5};
enum { arraylength = sizeof(array)/array[0] };
std::nth_element(array, array + 6, array + arraylength);
we finally have the array reordered such that:
- The element array[6] has the same value that it would have if the entire
array had been sorted
- The elements array[0] through array[5] are less than the elements in
the positions array[6] through array[11];
To my opinion this shows that the meaning of nth relates to a zero-
based index (provided that the actual iterator is an array) and not
to the usual ordinals description.
Daniel