Comment by peterhull90
14 days ago
You can put an extra 'level' in, to make both options return the same 0,1...9
def loop():
for number in range(10):
def outer(n):
def inner():
return n
return inner
yield outer(number)
Is there a neater way?
Yes. The usual way to do this is to bind them as defaults to an argument, for example:
This works because default arguments in python are evaluated exactly once, at function definition time. So it's a way of effectively copying the ``number`` out of the closure[1] and into the function definition.
[1] side note, closures in python are always late-binding, which is what causes the behavior in OP
I’ve never thought that leaking this type of implementation detail into the return value (and return type!) was a nice solution. I like the double closure better, and one can shorten it a bit with a lambda.
For those who prefer a functional style, functools.partial can also solve this problem.
(I use Python, and I like a lot of things about Python, but I don’t like its scoping rules at all, nor do I like the way that Python’s closures work. I would use a double lambda.)
Does this work for user defined classes (objects) as well?
Yes, and by the same mechanism and for the same reason.
I would have just done something like
``` def loop(): for number in range(10): fixed_number = number def inner(): return fixed_number yield inner ```
Correctly formatted (two spaces preceding each line, one blank line before the first code line, no extra lines needed between code lines):
The output:
It doesn't work because of the way Python's variables are scoped. Your fixed_number variable is still shared across all instances of inner. Python doesn't have any sort of block scoping like you seem to think it has.