Re: Type Functors

From:

Alan Johnson <awjcs@yahoo.com>

Newsgroups:

comp.lang.c++

Date:

Fri, 08 Jun 2007 09:57:48 -0700

Message-ID:

<VrmdnYcY298QF_TbnZ2dnUVZ_j-dnZ2d@comcast.com>

desktop wrote:

I have this class:

class test {
public:
    int getpp()
    {
        return pp;
    }

    void setpp(int i) {
        pp = i;
    }


private:
    int pp;
};

Now I would like to insert it into a set:

std::set<test, std::less<test> > my_set;

std::less<T> is the default for that template parameter. You do not
need to explicitly specify it (though it doesn't hurt). The only time
you'll use the second template parameter is if your type does not
implement operator< or you want an ordering other than that imposed by
operator<. Because you are implementing operator< below you can change
this to just:

std::set<test> my_set;

test t1;
my_set.insert(t1);

This does not work (missing a '<' operator) so I add the following to test:

class test {
public:
int getpp()

This function does not change the state of the class. It should be
declared 'const':
int getpp() const

    {
        return pp;
    }

    void setpp(int i) {
        pp = i;
    }

        bool operator <(test const& t) const {

        return (*this).getpp() < t.getpp();

Without the above change, you cannot do this. operator< has been
declared const, so you cannot call non-const member functions.

    }

private:
    int pp;
};

Which does not work either.

std::less implements something like:

bool operator() (T const& x, T const& y) const {
    return x < y;

}
So I don't quite see why I need to define '<' in 'test' when I supply
std::less<test> to my_set. It seems like double work. What is wrong in
the body of:

test::operator<(test const& t) const

--
Alan Johnson