Re: how to parallel sort?

From:
Kai-Uwe Bux <jkherciueh@gmx.net>
Newsgroups:
comp.lang.c++
Date:
Sun, 09 Mar 2008 14:31:36 -0400
Message-ID:
<fr1ae9$v6e$1@aioe.org>
nandor.sieben@gmail.com wrote:

You can easily use STL. Just create the resulting vector with indices
from 0 to n and sort it using as sort criteria the values of the other
vectors index. Give it some time to learn it, then come back if you
get stuck.

/Peter


This sounds like what I'd like to do but I don't know where to start
on changing the sort criteria. Do I need to write a function object
that has the other vector as extra input?


Yes.

I have a feeling this is
just a one liner sort command but it's beyond my knowledge of STL.


Consider:

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

template < typename Sequence >
class ForwardCompare {

  Sequence const & the_sequence;

public:

  ForwardCompare ( Sequence const & ref )
    : the_sequence ( ref )
  {}

  bool operator() ( typename Sequence::size_type lhs,
                    typename Sequence::size_type rhs ) const {
    return ( the_sequence[lhs] < the_sequence[rhs] );
  }

};

template < typename Sequence >
ForwardCompare< Sequence > make_ForwardCompare ( Sequence const & seq ) {
  return ( ForwardCompare< Sequence >( seq ) );
}

int main ( void ) {
  std::vector< int > value;
  value.push_back( 2 );
  value.push_back( 1 );
  value.push_back( 5 );
  value.push_back( 4 );
  std::vector< int > index;
  for ( int i = 0; i < value.size(); ++i ) {
    index.push_back( i );
  }
  std::sort( index.begin(), index.end(), make_ForwardCompare( value ) );
  std::copy( index.begin(), index.end(),
             std::ostream_iterator< int >( std::cout, " " ) );
  std::cout << '\n';
}

It would be a one-liner, if lambda allowed for:

  std::sort( index.begin(), index.end(), value[_1] < value[_2] );

However, I think that does not work (and as long as the subscript operator
is required to be a member function, I don't see a way to implement lambda
so that it would work).

Best

Kai-Uwe Bux

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