Re: [Boost-Users] lambda library bug --- endl?

From: Jaakko Jarvi (jajarvi_at_[hidden])
Date: 2002-09-13 13:21:40

> So if I explicitly list the template parameters, does it work?  Hmmm...
> 
> No, this doesn't work:
> 
> for_each(ell.begin(), ell.end(),
>     cout << _1 << endl<char, char_traits<char> >);
> 
> I was hopeful --- I'm pretty sure I have the syntax right.  I'm simply
> curious now: is it possible to make this work?

Explicit specialization should work, just tested under gcc3.2.

Here's another option, a helper function manip
that takes a stream and a manipulator and chooses the appropriate
instantiation of the endl template. Much nicer to just give the stream
object rather than the stream template args:

#include "boost/lambda/lambda.hpp"
#include <iostream>

namespace boost {
  namespace lambda {

    template <class CharT, class Traits>
    inline 
    std::basic_ostream<CharT, Traits>&
    (*manip(std::basic_ostream<CharT, Traits> &, 
            std::basic_ostream<CharT, Traits> &(*m)
              (std::basic_ostream<CharT, Traits> &)))
    (std::basic_ostream<CharT, Traits> &) 
    {
      return m;
    }

  }
}

int main () {

  using namespace std;
  using namespace boost::lambda;

  (cout << _1 << manip(cout, endl) << "Work!\n")("Seems to");
  return 0;
}

Cheers, Jaakko

> 
> Ken
> --
> Ken Yarnall
> Dept. of Mathematical Sciences               Lebanon Valley College
> Assoc. Professor of Math and CS       [phone] yarnall_at_[hidden]
> 
> 
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-- 
--
-- Jaakko Järvi                       email: jajarvi_at_[hidden]
-- Post Doctoral Fellow               phone: [phone]
-- Pervasive Technology Labs          fax:   [phone]
-- Indiana University, Bloomington