Seemingly odd 'is' comparison.
Boris Borcic
bborcic at gmail.com
Tue Feb 19 14:47:06 CET 2008
Arnaud Delobelle wrote:
> On Feb 13, 10:19 pm, Tobiah <t... at tobiah.org> wrote:
>>>>> print float(3.0) is float(3.0)
>> True
>>>>> print float(3.0 * 1.0) is float(3.0)
>> False
>
> [You don't need to wrap your floats in float()]
>
>>>> def f():
> ... return 3.0 is 3.0, 3.0*1.0 is 3.0
> ...
>>>> f()
> (True, False)
>>>> import dis
>>>> dis.dis(f)
> 2 0 LOAD_CONST 1 (3.0)
> 3 LOAD_CONST 1 (3.0)
> 6 COMPARE_OP 8 (is)
> 9 LOAD_CONST 3 (3.0)
> 12 LOAD_CONST 1 (3.0)
> 15 COMPARE_OP 8 (is)
> 18 BUILD_TUPLE 2
> 21 RETURN_VALUE
>
> As you can see when "3.0 is 3.0" is evaluated the same float object is
> put on the stack twice so the 'is' comparison is True (LOAD_CONST 1 /
> LOAD_CONST 1 / COMPARE_OP 8).
>
> Whereas when "3.0*1.0 is 3.0" is evaluated, *two* different float
> objects are put on the stack and compared (LOAD_CONST 3 / LOAD_CONST
> 1 / COMPARE_OP 8). Therefore the result is False.
Looks good, but doesn't pass the sanity check ;) Consider
>>> def f():
return 3 is 3, 3*1 is 3
>>> import dis
>>> dis.dis(f)
2 0 LOAD_CONST 1 (3)
3 LOAD_CONST 1 (3)
6 COMPARE_OP 8 (is)
9 LOAD_CONST 3 (3)
12 LOAD_CONST 1 (3)
15 COMPARE_OP 8 (is)
18 BUILD_TUPLE 2
21 RETURN_VALUE
>>> f()
(True, True)
Cheers, BB
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