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

It is not silently ignored. It is used as an argument to construct StopIteration. See The Python Language Reference:

https://docs.python.org/3/reference/simple_stmts.html#the-return-statement

Apologies if I wasn't clear. I understand that

def f():
    yield 1
    return

is valid python. What I'm saying, if you follow the link, is that

def f():
    yield 1
    return 2  # not the argument

should not be considered valid python according to PEP 255. This is implemented in python 2 but not in python 3.

Whoops I understand. Reclosed.

See also What’s New In Python 3.3:

https://docs.python.org/3/whatsnew/3.3.html#pep-380-syntax-for-delegating-to-a-subgenerator

And PEP 380 itself.

nosy: + serhiy.storchaka
messages: + msg316913

resolution: not a bug
stage: resolved