Serhiy Storchaka added the comment: What if the third-party parser don't use prepare_input_source()? It can use more efficient way if pass just a file name.
Wouldn't be better to move your code into the parser's method parse()? If a file is opened inside the parse() method and is not exposed outside, that method should close it. ---------- components: +Library (Lib) -Tests nosy: +eli.bendersky, scoder, serhiy.storchaka stage: -> patch review type: -> behavior _______________________________________ Python tracker <rep...@bugs.python.org> <http://bugs.python.org/issue30264> _______________________________________ _______________________________________________ Python-bugs-list mailing list Unsubscribe: https://mail.python.org/mailman/options/python-bugs-list/archive%40mail-archive.com
