Question

Let’s say I have a Python function f and fhelp. fhelp is designed to call itself recursively. f should not be called recursively. Is there a way for f to determine if it has been called recursively?

Answer 1

Use the traceback module for this:

>>> import traceback
>>> def f(depth=0):
...     print depth, traceback.print_stack()
...     if depth < 2:
...         f(depth + 1)
...
>>> f()
0  File "<stdin>", line 1, in <module>
  File "<stdin>", line 2, in f
 None
1  File "<stdin>", line 1, in <module>
  File "<stdin>", line 4, in f
  File "<stdin>", line 2, in f
 None
2  File "<stdin>", line 1, in <module>
  File "<stdin>", line 4, in f
  File "<stdin>", line 4, in f
  File "<stdin>", line 2, in f
 None

So, if any entry in the stack indicates that the code was called from f, the call was (in)directly recursive. The traceback.extract_stack method gives you an easy access to this data. The if len(l[2] ... statement in the example below simply counts the number of exact matches of the name of the function. To make it even prettier (thanks to agf for the idea), you could make it into a decorator:

>>> def norecurse(f):
...     def func(*args, **kwargs):
...         if len([l[2] for l in traceback.extract_stack() if l[2] == f.__name__]) > 0:
...             raise Exception('Recursed')
...         return f(*args, **kwargs)
...     return func
...
>>> @norecurse
... def foo(depth=0):
...     print depth
...     foo(depth + 1)
...
>>> foo()
0
Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
  File "<stdin>", line 5, in func
  File "<stdin>", line 4, in foo
  File "<stdin>", line 5, in func
Exception: Recursed