Is the shortcircuit behaviour of Python's any/all explicit?
PythonShort CircuitingLanguage SpecificationsPython Problem Overview
Prompted by the discussion here
The docs suggest some equivalent code for the behaviour of all
and any
Should the behaviour of the equivalent code be considered part of the definition, or can an implementation implement them in a non-shortcircuit manner?
Here is the relevant excerpt from cpython/Lib/test/test_builtin.py
def test_all(self):
self.assertEqual(all([2, 4, 6]), True)
self.assertEqual(all([2, None, 6]), False)
self.assertRaises(RuntimeError, all, [2, TestFailingBool(), 6])
self.assertRaises(RuntimeError, all, TestFailingIter())
self.assertRaises(TypeError, all, 10) # Non-iterable
self.assertRaises(TypeError, all) # No args
self.assertRaises(TypeError, all, [2, 4, 6], []) # Too many args
self.assertEqual(all([]), True) # Empty iterator
S = [50, 60]
self.assertEqual(all(x > 42 for x in S), True)
S = [50, 40, 60]
self.assertEqual(all(x > 42 for x in S), False)
def test_any(self):
self.assertEqual(any([None, None, None]), False)
self.assertEqual(any([None, 4, None]), True)
self.assertRaises(RuntimeError, any, [None, TestFailingBool(), 6])
self.assertRaises(RuntimeError, all, TestFailingIter())
self.assertRaises(TypeError, any, 10) # Non-iterable
self.assertRaises(TypeError, any) # No args
self.assertRaises(TypeError, any, [2, 4, 6], []) # Too many args
self.assertEqual(any([]), False) # Empty iterator
S = [40, 60, 30]
self.assertEqual(any(x > 42 for x in S), True)
S = [10, 20, 30]
self.assertEqual(any(x > 42 for x in S), False)
It doesn't do anything to enforce the shortcircuit behaviour
Python Solutions
Solution 1 - Python
The behaviour is guaranteed. I've contributed a patch, which was accepted and merged recently, so if you grab the latest sources you will see that the short-circuiting behaviour is now explicitly enforced.
git clone https://github.com/python/cpython.git
grep Short-circuit cpython/Lib/test/test_builtin.py
Solution 2 - Python
The docs say
>"Return True if any element of the iterable is true. If the iterable is empty, return False. EQUIVALENT TO:" (emphasis mine) ...
def any(iterable):
for element in iterable:
if element:
return True
return False
If any
didn't short circuit, it wouldn't be EQUIVALENT to the posted code since the posted code clearly short circuits. You could consume more of a generator than you want to for example. In light of that, I say that the short circuiting behavior is guaranteed.
The exact same argument could be made for all
.
Solution 3 - Python
In case you landed here looking for "do any/all always always short-circuit?"
They do, but there is a gotcha: using a list comprehension can make it seem like you are overriding the short-circuiting behavior:
def hi():
print('hi')
return True
>>> any(hi() for num in [1, 2, 3, 4])
hi
>>> any([hi() for num in [1, 2, 3, 4]])
hi
hi
hi
hi
The list comprehension executes before any() does.
(Note: This does not answer the OP's very different question. This is the only stackoverflow page that comes up for me when searching "any all short circuit python.")
Solution 4 - Python
It HAS to short circuit, since it could be given an unbound iterable. If it did not short circuit then this would never terminate:
any(x == 10 for x in itertools.count())