New submission from FHTMitchell <fergus....@gmail.com>: In python 2.7 if you run the following code you get an error (as you would expect)
Python 2.7.14 | packaged by conda-forge | (default, Dec 25 2017, 01:17:32) [MSC v.1500 64 bit (AMD64)] on win32 Type "help", "copyright", "credits" or "license" for more information. >>> def f(): ... yield 1 ... return 2 ... File "<stdin>", line 3 SyntaxError: 'return' with argument inside generator However, in python 3.6 the error is silently ignored Python 3.6.4 | packaged by conda-forge | (default, Dec 24 2017, 10:11:43) [MSC v.1900 64 bit (AMD64)] on win32 Type "help", "copyright", "credits" or "license" for more information. >>> def f(): ... yield 1 ... return 2 ... >>> for i in f(): ... print(i) ... 1 and still is in 3.7 Python 3.7.0b2 (v3.7.0b2:b0ef5c979b, Feb 28 2018, 02:24:20) [MSC v.1912 64 bit (AMD64)] on win32 Type "help", "copyright", "credits" or "license" for more information. >>> def f(): ... yield 1 ... return 2 ... >>> for i in f(): ... print(i) ... 1 This is a source of confusion https://stackoverflow.com/questions/47831240/why-is-no-value-returned-from-my-generator/ especially since the PEP says it is disallowed: https://www.python.org/dev/peps/pep-0255/#then-why-not-allow-an-expression-on-return-too ---------- components: Interpreter Core messages: 316912 nosy: FHTMitchell priority: normal severity: normal status: open title: No SyntaxError raised for `return` with argument inside generator type: behavior versions: Python 3.6, Python 3.7 _______________________________________ Python tracker <rep...@bugs.python.org> <https://bugs.python.org/issue33555> _______________________________________ _______________________________________________ Python-bugs-list mailing list Unsubscribe: https://mail.python.org/mailman/options/python-bugs-list/archive%40mail-archive.com
