Issue 33555: No SyntaxError raised for return with argument inside generator (original) (raw)

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 "", 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