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 (717)867-6085 yarnall@lvc.edu
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-- -- -- Jaakko Järvi email: jajarvi@cs.indiana.edu -- Post Doctoral Fellow phone: +1 (812) 855-3608 -- Pervasive Technology Labs fax: +1 (812) 855-4829 -- Indiana University, Bloomington