Hi Ken,
for_each(ell.begin(), ell.end(), cout << _1 << endl);
Any suggestions? I am aware that I can replace endl by '\n' and get a compilable program. I worry that this problem extends to other manipulators...
It's not a bug, just an unfortunate restriction. This is an extract form Lambda wiki page: http://www.crystalclearsoftware.com/cgi-bin/boost_wiki/wiki.pl?Lambda Example: endl is not now allowed in std::cout statements in a Lambda expression. '\n' has to be used instead. -- This is because the old style streams were not templates and the new style ones are. Also, endl has been changed from a non-template function to a template function and one cannot take the address of a function template (withtout instantiated it at the same time. For example, in _1 << endl, the library does not know what the type of the stream that later is substituted for _1 will be. -- Jaakko Jarvi Jaakko