Seemingly odd 'is' comparison.
Arnaud Delobelle
arnodel at googlemail.com
Tue Feb 19 12:54:14 EST 2008
On Feb 19, 1:47 pm, Boris Borcic <bbor... at gmail.com> wrote:
> 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
Looks good, but doesn't pass the sanity check ;) Consider:
>>> def f():
... return 1000000 is 1000000, 1000000*1 is 1000000
...
>>> f()
(True, False)
>>> dis.dis(f)
2 0 LOAD_CONST 1 (1000000)
3 LOAD_CONST 1 (1000000)
6 COMPARE_OP 8 (is)
9 LOAD_CONST 3 (1000000)
12 LOAD_CONST 1 (1000000)
15 COMPARE_OP 8 (is)
18 BUILD_TUPLE 2
21 RETURN_VALUE
Small integers are special because they are interned, that's why you
got True for the second comparison. With a larger integer that
doesn't happen.
--
Arnaud
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