The following compiles: (cout << _1 << "\n")(x); but this doesn't: (cout << _1 << endl)(x); (gcc 3.2.2) I can't figure out how to use an io manipulator within a lambda function. Firstly, how do I do it? (e.g. suppose I want to use flush, for which there is no workaround like using "\n"). Secondly, a suggestion: it would be nice if support was built in for the io manipulators. - Chuck