Welcome to OStack Knowledge Sharing Community for programmer and developer-Open, Learning and Share
Welcome To Ask or Share your Answers For Others

Categories

0 votes
309 views
in Technique[技术] by (71.8m points)

short circuiting - Is the shortcircuit behaviour of Python's any/all explicit?

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

See Question&Answers more detail:os

与恶龙缠斗过久,自身亦成为恶龙;凝视深渊过久,深渊将回以凝视…
Welcome To Ask or Share your Answers For Others

1 Answer

0 votes
by (71.8m points)

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

与恶龙缠斗过久,自身亦成为恶龙;凝视深渊过久,深渊将回以凝视…
Welcome to OStack Knowledge Sharing Community for programmer and developer-Open, Learning and Share
Click Here to Ask a Question

2.1m questions

2.1m answers

60 comments

57.0k users

...