Write recursive function that acts like sys.getrecursionlimit

Question

How to write a recursive function that acts like sys.getrecursionlimit, thus getting the recursion limit without importing any libraries?

def recurse(n):
m = 0
def recurse2(n):
    nonlocal m 
    m = m+1
    n-recurse2(n-1)
try: 
    recurse2(n)
except RecursionError:
    print(m)

This is what I've attempted so far.

Answer 1

Its easiest to pass a counter down to each function rather than try to access variables out of the function scope:

def get_limit():
    try:
        return 1 + get_limit()
    except RecursionError:
        return 2

which gives 1000 for me - just like sys.getrecursionlimit().

---

why?

So on every call of the function, the function adds one to the result of how many more recursive calls can be made. The answer to this question "how many [more] recursive calls can be made?" is simply answered by the function itself, so we return 1 + get_limit() since we were called so we must return one more.

Finally, we must define a base case which will lie at the bottom of the tree and handle when the answer to "how many [more] recursive calls can be made?" is RecursionError, i.e. "no more". In this case, the true answer is 1, but since the actual function will be called once, we should return 2 to account for the call at the top of the tree/stack so that our result is, in my case, 1000 rather than the 999 that it would be if we returned 1.

Answer 2

You can use try/except inside the recursive function to be able to return the depth counter:

def _get_recursion_limit(n = 1):
  try:
     return _get_recursion_limit(n+1)
  except RecursionError:
     return n+1 #account for last attempt with increment

import sys
print(_get_recursion_limit())
print(sys.getrecursionlimit())

Output:

1000
1000

On my machine, both results are 1000.