Re: [Boost-users] analogy to mem_fun_ref to member variables

From: Stuart Dootson (stuart.dootson_at_[hidden])
Date: 2005-01-27 03:55:24

On Wed, 26 Jan 2005 17:21:43 -0700, Chris Goller
<cgoller_at_[hidden]> wrote:
> I have a struct that contains two values.
> 
> struct box
> {
>     double width;
>     double length;
> }
> 
> vector<box> boxes;
> 
> I want to accumlate all the widths of the boxes.  The only solution I've
> found is to create a member function get_width that
> will return the width for use in mem_fun_ref
> 
> eg:
> 
> accumulate(make_transform_iterator(boxes.begin(),
> mem_fun_ref(&boxes::get_width)),
> make_transform_iterator(boxes.end(), mem_fun_ref(&boxes::get_width)),
> 0.0,
> plus<double>()));
> 
> Is there a boost way of not using mem_fun_ref, but rather returns the
> public member variable.  Something that could
> be named mem_var_ref that would look like this:
> 
> accumulate(make_transform_iterator(boxes.begin(),
> mem_var_ref(&boxes::width)),
> make_transform_iterator(boxes.end(), mem_fun_ref(&boxes::width)),
> 0.0,
> plus<double>()));
> 
> This way I wouldn't need to add extra get logic to a structure that
> doesn't really need it othewise.
> 
> Thanks,
> 
> Chris
>

Boost.Bind will do what you want, I think - check out this example
which sums the x_ members of an array of structs of type A as shown
below:

#include <boost/bind.hpp>
#include <iostream>
#include <numeric>
#include <functional>
#include <algorithm>

struct A
{
   int x_;
   int y_;
};

int main(int, char**)
{
   A o[3] = { {1, 2}, {2, 3}, {3, 4} };

   int sum = std::accumulate(&o[0], &o[3], 0,
boost::bind(std::plus<int>(), boost::bind(&A::x_, _2), _1));
   std::cout << sum << "\n"; // Gives the expected answer - 6!!
}

Rather than use a transform iterator, I've used function composition
to transform the argument of type A to an integer (by accessing the x_
member), before applying the std::plus object.

HTH

Stuart Dootson