"Permanent" std::setw

Question

Is there any way how to set std::setw manipulator (or its function width) permanently? Look at this:

#include <iostream>
#include <iomanip>
#include <algorithm>
#include <iterator>

int main( void )
{
  int array[] = { 1, 2, 4, 8, 16, 32, 64, 128, 256 };
  std::cout.fill( '0' );
  std::cout.flags( std::ios::hex );
  std::cout.width( 3 );

  std::copy( &array[0], &array[9], std::ostream_iterator<int>( std::cout, " " ) );

  std::cout << std::endl;

  for( int i = 0; i < 9; i++ )
  {
    std::cout.width( 3 );
    std::cout << array[i] << " ";
  }
  std::cout << std::endl;
}

After run, I see:

001 2 4 8 10 20 40 80 100

001 002 004 008 010 020 040 080 100

I.e. every manipulator holds its place except the setw/width which must be set for every entry. Is there any elegant way how to use std::copy (or something else) along with setw? And by elegant I certainly don't mean creating own functor or function for writing stuff into std::cout.

Answer 1

Well, it's not possible. No way to make it call .width each time again. But you can use boost, of course:

#include <boost/function_output_iterator.hpp>
#include <boost/lambda/lambda.hpp>
#include <algorithm>
#include <iostream>
#include <iomanip>

int main() {
    using namespace boost::lambda;
    int a[] = { 1, 2, 3, 4 };
    std::copy(a, a + 4, 
        boost::make_function_output_iterator( 
              var(std::cout) << std::setw(3) << _1)
        );
}

It does create its own functor, but it happens behind the scene :)