Re: Type Functors

From:

Alan Johnson <awjcs@yahoo.com>

Newsgroups:

comp.lang.c++

Date:

Fri, 08 Jun 2007 13:20:45 -0700

Message-ID:

<f4cdmt$ce9$1@aioe.org>

desktop wrote:

Alan Johnson wrote:

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<.

The problem seems to occur when I do an insert:

test t1;
my_set.insert(t1);

as long as I don't insert I can make a set with or without specifying
std::less<test> and without implementing the '<' in test.

When doing an insert I must implement the '<' operator in test and it
makes no difference if I specify std::less<test> or not.

But I am not sure what to put in the body of '<'. Currently I have:

    bool operator <(test const& t) const {
        return (*this).getpp() < t.getpp();
    }

but that was just to fill in the body.

One more, you cannot call a non-const member function from within a
const member function. To implement operator< the way you are trying to
do it, you need to make getpp const.

#include <set>

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

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

     bool operator<(const test & t) const
     {
         return getpp() < t.getpp();
     }

private:
     int pp;
};

int main()
{
     std::set<test> my_set;
     test t1;
     my_set.insert(t1);
}

--
Alan Johnson