Re: Providing a no-overhead way for a contained class to access its container?

From:

Jerry Coffin <jcoffin@taeus.com>

Newsgroups:

comp.lang.c++

Date:

Tue, 17 Jun 2008 00:00:11 -0600

Message-ID:

<MPG.22c1009236fab9d7989d52@news.sunsite.dk>

In article <5dd80816-49c4-442d-a25c-076b6c260305
@y21g2000hsf.googlegroups.com>, PeteOlcott@gmail.com says...

[ ... ]

The only reason that I even need to have the contained class is so
that I can overload the operator<() on it, and thus use std::sort().

You can use std::sort without having an overloaded operator< for the
type. You can create the comparison as either a function or a functor,
and then pass it as the third parameter to std::sort. For example:

#include <vector>
#include <algorithm>
#include <iostream>

class X {
    int y;
public:
    X(int a=0) : y(a) {}
    operator int() const { return y; }
};

struct less {
    bool operator()(X const &a, X const &b) {
        return (int)a < (int)b;
    }
};

int main() {
    std::vector<X> x;

    for (int i=0; i<20; i++)
        x.push_back(rand());

    std::sort(x.begin(), x.end(), less());
    std::copy(x.begin(), x.end(),
        std::ostream_iterator<int>(std::cout, "\n"));
    return 0;
}

--
    Later,
    Jerry.

The universe is a figment of its own imagination.