question about generators
jonathan at onegoodidea.com
Wed Aug 14 22:36:38 CEST 2002
On 14/8/2002 16:38, in article yu993cthtpqr.fsf at europa.research.att.com,
"Andrew Koenig" <ark at research.att.com> wrote:
> I had a function along the following lines:
> def f():
> if <condition>:
> print <something>
> <do something>
> <do something else>
> and I wanted to turn it into a generator instead of having it
> generate output directly. My first try was to change "print"
> to "yield", and that failed horribly.
I'm confused, the pattern you show above will only ever 'print <something>'
once. There are only two paths through the function, one prints, the other
calls itself recursively. So the function can only work down a bit through
some recursive calls doing <something>, then when the condition is true it
will print something, bubble back up through all the calls doing <something
else>, and exit.
So if it can only return one thing, you don't need a generator. So I'm
guessing there's something crucial missing from the pattern shown.
I was going to suggest analysing the function and turning it into a loop
instead of recursion, but I can't suggest how to do that without knowing a
bit more about how it was meant to work.
That said, I usually just do what you did and put a 'for x in f(): yield x'
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