In Python, will memory for variables in a recursive function be freed if they're not used after the recursive call?

Question

Below is some code written iteratively with mild resemblance to the DBSCAN algorithm:

rw = 100
for r in range(rw):
    neighs = retrieve_neighs(r)
    # rest of body

If I rewrite it using recursion, I might write:

rw = 100
def foo(r):
    if r < rw:
        neighs = retrieve_neighs(r)
        # rest of body
        foo(r + 1)
foo(0)

I expect retrieve_neighs to return a list (or some collection) of points - this opens up the possibility that an entire file of data could be loaded, right? Is there any recognition that neighs is not used after

foo(r + 1)

and so the memory neighs refers to is freed?

You can delete your neighs object once you are done with performing some operation on it. And if your are returning a list then you can create a generator object instead of whole list. — Tanveer Alam
– Tanveer Alam, Commented Dec 27, 2014 at 4:59

lvc · Accepted Answer · 2014-12-27 05:07:01Z

5

What you're describing is a major effect of tail call optimization, which Python explicitly doesn't do because Guido feels that it isn't worth the tradeoffs, especially in terms of debugging information. If the recursive version of the algorithm is clearer, you might consider explicitly doing del neighs once you no longer need it - this has the save effect as the variable falling out of scope, potentially making it eligible for garbage collection.

edited Dec 27, 2014 at 5:07

answered Dec 27, 2014 at 5:00

lvc

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