Python – Fixing SyntaxError: ‘return’ with argument inside generator

This error is telling you that when you use a yield inside of a function making it a generator, you can only use return with no arguments.

This makes sense because return raises a StopIteration error in a generator and it would be quite unexpected to be getting values back with that.

Here is how you can accomplish what you need to do though:

Say you want to do this:

def f():
    for i in range(2):
        yield i
    return i+2

You can do this instead and probably get a result close enough to what you want:

def f():
    for i in range(2):
        yield i
    yield i+2
    return

Posted by Greg Pinero (Primary Searcher) on Mar 26th, 2008 and is filed under Python.

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2 Responses to “Python – Fixing SyntaxError: ‘return’ with argument inside generator”

  1. Jonathan Ballet says:
    March 26th, 2008 at 9:10 am

    You can even remove the “return” statement in the second example, it is not needed at all.

  2. Greg Pinero (Primary Searcher) says:
    March 26th, 2008 at 9:45 am

    Jonathan, you are correct. I wanted to include it to show how a return in a generator would look in case you’re doing something more complex. For example wanting to return from inside a loop at a certain point.