Re: Member function as parameter of STL algorithm

From:
 "int2str@gmail.com" <int2str@gmail.com>
Newsgroups:
comp.lang.c++
Date:
Wed, 18 Jul 2007 17:22:26 -0700
Message-ID:
<1184804546.139696.279990@z24g2000prh.googlegroups.com>
On Jul 17, 7:42 am, many_years_after <shua...@gmail.com> wrote:

Hi, cppers:

     I am studying cpp recently. As is said, member funciton can be as
one parameter of stl algorithm. BUT when I pass member function whose
parameter is instance of the class, it doesn't work. For example:


It won't work. std::sort() either needs a predicate function
(functor), or it simply needs the < (less than) operator to work
correctly.

The latter case is the easiest. Let's make your code work...

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

using namespace std;

class Point
{
public:
    int x; int y;
    Point(int xx, int yy)
    {
        x = xx;
        y = yy;
    }
    friend ostream& operator<<(ostream& out,const Point& p)
    {
        out << p.x << " "<< p.y << endl;
        return out;
    }
    bool LargeThan(const Point& p)


bool LargerThan(const Point& p) const // const is IMPORTANT
here...

    {

        return (x > p.x)|| ((x==p.x) && (y > p.y));
    }
    void print()
    {
        cout << x << " " << y;
    }
    void printWithPre(const char* s)
    {
        cout << s << " " << x;
    }


    bool operator< (const Point& p ) const
    {
        return !LargerThan(p);
    }

    };

int main()
{
    vector<Point> vec;

    for (int i = 0; i < 10; i++)
        vec.push_back(Point(i, i));
    for_each(
        vec.begin(),
        vec.end(),
        bind2nd(mem_fun_ref(&Point::printWithPre),"hello:")
        ); //OK

    for_each(
                vec.begin(),
        vec.end(),
        mem_fun_ref(&Point::print)); // OK
   sort(vec.begin(), vec.end(), mem_fun_ref(&Point::LargeThan)); //
ERROR WHEN COMPILING


Now that the operator is defined, simply use:

    sort(vec.begin(), vec.end());

    return 0;

}

I don't know why it does not work. What's the reasong?

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