Re: [Boost-users] [lambda] Confusion!

22 Mar 2008


      On Fri, Mar 21, 2008 at 10:16 PM, Steven Watanabe 
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 
#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, 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.