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Subject: [Boost-users] [iterator] std::sort on zip_iterator
From: Jeremiah Willcock (jewillco_at_[hidden])
Date: 2009-05-02 13:11:36
I would like to sort two separate vectors of numbers, treating
corresponding elements of the two vectors as tuples and sorting the
tuples. In particular, I am only comparing the first elements of the
tuples; that may or may not be relevant to my problem. An example program
is:
#include <vector>
#include <algorithm>
#include <iostream>
#include <boost/tuple/tuple.hpp>
#include <boost/iterator/zip_iterator.hpp>
struct compare_first_elements_in_tuples {
template <typename Tuple>
bool operator()(const Tuple& a, const Tuple& b) const {
return (a.template get<0>()) < (b.template get<0>());
}
};
int main(int, char**) {
size_t a_data[5] = {0, 2, 1, 2, 4};
size_t b_data[5] = {1, 1, 5, 3, 7};
std::vector<size_t> a(a_data, a_data + 5);
std::vector<size_t> b(b_data, b_data + 5);
std::cout << "Before:";
for (size_t i = 0; i < a.size(); ++i) {
std::cout << " (" << a[i] << ", " << b[i] << ")";
}
std::cout << std::endl;
typedef std::vector<size_t>::iterator vec_iter;
typedef boost::zip_iterator<boost::tuple<vec_iter, vec_iter> > zip_iter;
std::sort(zip_iter(boost::make_tuple(a.begin(), b.begin())),
zip_iter(boost::make_tuple(a.end(), b.end())),
compare_first_elements_in_tuples());
std::cout << "After:";
for (size_t i = 0; i < a.size(); ++i) {
std::cout << " (" << a[i] << ", " << b[i] << ")";
}
std::cout << std::endl;
return 0;
}
On gcc 4.0.1, the sort operation results in the element (2, 1) (a pair of
corresponding elements in a and b) being duplicated, while the element (1,
5) does not appear in the result at all:
Before: (0, 1) (2, 1) (1, 5) (2, 3) (4, 7)
After: (0, 1) (2, 1) (2, 1) (2, 3) (4, 7)
I suspect the problem is related to the fact that zip_iterator does not
return a "real" reference when dereferenced. Am I mistaken about this?
Is the code above supposed to work? If not, is there an alternative way
to sort one sequence while swapping elements in another in a corresponding
way? I am not able to join the two input sequences into a single sequence
of pairs because of other factors in the use case for this code. Thank
you for your help.
-- Jeremiah Willcock
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