Unable to return a generator instance. why?

Question

I am using python 3.5. When I tried to return a generator function instance and i am getting a StopIteration error. Why?

here is my code:

>>> def gen(start, end):
... '''generator function similar to range function'''
...    while start <= end:
...       yield start
...       start += 1
...
>>> def check(ingen, flag=None):
...    if flag:
...        for n in ingen:
...            yield n*2
...    else:
...        return ingen
...
>>> # Trigger else clause in check function
>>> a = check(gen(1,3))
>>> next(a)
Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
StopIteration: <generator object gen at 0x7f37dc46e828>

It looks like the generator is somehow exhausted before the else clause is returns the generator.

It works fine with this function:

>>> def check_v2(ingen):
...     return ingen
...
>>> b = check_v2(gen(1, 3))
>>> next(b)
1
>>> next(b)
2
>>> next(b)
3

@wroniasty: From what I've just read, it fine only in Python 3.3+ — martineau, Aug 13 '16 at 03:22

score 3 · Accepted Answer · edited May 23 '17 at 12:33

In Python, if yield is present in a function, then Python treats it as a generator. In a generator, any return will raise StopIteration with the returned value. This is a new feature in Python 3.3: see PEP 380 and here. check_v2 works because it doesn't contain a yield and is therefore a normal function.

There are two ways to accomplish what you want:

Change the return to a yield in check.

Have the caller trap StopIteration, as shown below

try:
    next(a)
except StopIteration as ex:
    print(ex.value)

score 2 · Answer 2 · answered Aug 12 '16 at 15:49

2

When a generator hits its return statement (explicit or not) it raises StopIteration. So when you return ingen you end the iteration.

check_v2 is not a generator, since it does not contain the yield statement, that's why it works.

answered Aug 12 '16 at 15:49

wroniasty

7,884
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score 1 · Answer 3 · answered Aug 12 '16 at 15:52

1

As others have said, if you return from a generator, it means the the generator has stopped yielding items, which raises a StopIteration whatever you return.

This means that check actually returns an empty iterator.

If you want to return the results of another generator, you can use yield from :

def check(ingen, flag=None):
    if flag:
        for n in ingen:
            yield n*2
    else:
        yield from ingen

answered Aug 12 '16 at 15:52

pistache

5,782
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1

Thank u for opening my eyes to the `yield from` syntax. +1 – Aug 12 '16 at 16:02
1

interestingly it's also with `yield from` that coroutines are implemented (were, actually, now it's a native opcode) – pistache Aug 12 '16 at 16:12

score 0 · Answer 4 · answered Aug 12 '16 at 15:56

See PEP 380:

In a generator, the statement
return value
is semantically equivalent to
raise StopIteration(value)
except that, as currently, the exception cannot be caught by except clauses within the returning generator.

This is a new feature in Python 3.3.

Unable to return a generator instance. why?

4 Answers4