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

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);

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 inline std::basic_ostream& (*manip(std::basic_ostream &, std::basic_ostream &(*m) (std::basic_ostream &))) (std::basic_ostream &) { 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