On Fri, Mar 21, 2008 at 10:16 PM, Steven Watanabe <watanabesj@gmail.com> wrote:
AMDG

Robert Dailey wrote:
> Hi,
>
> I have defined a global overloaded operator<<. I want to use this
> overloaded operator in std::for_each(), but it must be properly
> deduced since there are many global overloads for this operator. For
> example:
> <snip>
>
>
> What am I doing wrong? Do I need to be using boost bind instead? If
> so, how can I use boost.bind to map a global operator? Will it
> correctly deduce which overload to use?

Unfortunately, Boost.Lambda does not handle user defined operators
automatically.
If you don't do this Lambda will try to return a Stream by value which
apparently
triggers SFINAE when Stream is abstract.

#include <iostream>
#include <boost/lambda/lambda.hpp>
#include <vector>
#include <algorithm>

class Stream
{
public:
   virtual void abstract() = 0;
   // This is an abstract class (cannot instantiate)
   void write(char ch) {
       std::cout << ch;
   }
};

class MyStream : public Stream {
   void abstract() {}
};

Stream& operator<< ( Stream& stream, char data )
{
   stream.write( data );
   return stream;
}

namespace boost {
namespace lambda {

template<>
struct plain_return_type_2<bitwise_action<leftshift_action>, Stream, char>
{
 typedef Stream& type;
};

}
}

Stream& operator<< ( Stream& stream, std::vector<char>& buffer )
{
   using namespace boost::lambda;
   std::for_each( buffer.begin(), buffer.end(), var( stream ) << _1 );
   return stream;
}

int main() {
   MyStream s;
   std::string str("test1");
   std::vector<char> b(str.begin(), str.end());
   s << b;
}

See http://www.boost.org/doc/html/lambda/extending.html

Will the plain_return_type_2 specialization also work for classes derived from Stream? Or do I need to specialize that structure for each and every single class, regardless of inheritance? Thanks.