Steven D'Aprano <steve+pyt...@pearwood.info> added the comment:
Tim is correct, the behaviour is right, however the docs could be clearer.
I think what you are missing is that ``or`` and ``and`` are short-circuiting
operators. So in the expression
9 or (anything)
the "anything" expression never gets evaluated because 9 is a truthy value. You
might get a better example of what is happening if you disassemble the
byte-code:
py> from dis import dis
py> dis(compile("9 or 7 < 'str'", '', 'eval'))
1 0 LOAD_CONST 0 (9)
3 JUMP_IF_TRUE_OR_POP 15
6 LOAD_CONST 1 (7)
9 LOAD_CONST 2 ('str')
12 COMPARE_OP 0 (<)
>> 15 RETURN_VALUE
Another way to see the order of evaluation is if we make the left hand operand
falsey:
py> print("first") or print("second") < 'str'
first
second
Traceback (most recent call last):
File "<stdin>", line 1, in <module>
TypeError: unorderable types: NoneType() < str()
In this case, the order of operations is:
- evaluate ``print("first")`` (returns None)
- since None is falsey, evaluate ``print("second") < 'str'``
- which prints the word "second", then raises an exception.
We can get rid of the exception:
py> print("first") or str(print("second")) < 'str'
first
second
True
I can see how the docs are a little bit confusing if you don't remember that
``or`` and ``and`` are short-circuiting operators. If you would like to propose
an improvement to the docs, please suggest something. But the behaviour is
correct and is not a bug.
https://docs.python.org/3/reference/expressions.html#operator-precedence
----------
nosy: +steven.daprano
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<https://bugs.python.org/issue38060>
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