using "private" parameters as static storage?
Peter Otten
__peter__ at web.de
Fri Nov 14 03:49:04 EST 2008
Steven D'Aprano wrote:
> On Thu, 13 Nov 2008 19:52:13 -0700, Joe Strout wrote:
>
>> Pity there isn't a way for a function to get a reference to itself
>> except by name. Still, when you rename a method, you're going to have
>> to update all the callers anyway -- updating a couple of extra
>> references within the method is not that much extra effort.
>
> I have come across a situation where this was a problem.
>
> I was writing a bunch of different versions of the classic factorial and
> Fibonacci functions, some iterative, some recursive, some with caches,
> and some without. What I wanted to do was do something like this (for
> factorial):
>
>
> def iterative_without_cache(n):
> product = 1
> for i in xrange(1, n+1):
> product *= i
> return product
>
> iterative_with_cache = decorate_with_cache(iterative_without_cache)
>
> which works. But it doesn't work for the recursive version:
>
> def recursive_without_cache(n):
> if n <= 1:
> return 1
> else:
> return n*recursive_without_cache(n-1)
>
> recursive_with_cache = decorate_with_cache(recursive_without_cache)
>
> for obvious reasons. Solution: the copy-and-paste anti-pattern. Good
> enough for test code, not for real work.
>
> Now multiply the above by about a dozen slightly different recursive
> versions of the Fibonacci function, and you will see why it was a problem.
You can have Python do the copying for you:
def make_recursive(deco):
@deco
def recursive(n):
if n <= 1:
return 1
return n * recursive(n-1)
return recursive
def cached(f, _cache={}):
def g(*args):
try:
result = _cache[args]
except KeyError:
result = _cache[args] = f(*args)
return result
return g
r1 = make_recursive(lambda f: f)
r2 = make_recursive(cached)
Peter
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