Immutable and Mutable Types
Bernard Lim
I'm reading the Python Reference Manual in order to gain a better understanding
of Python under the hood.
On the last paragraph of 3.1, there is a statement on immutable and mutable
types as such:
<paraphrase>
Depending on implementation, for immutable types, operations that compute
new values may or may not actually return a reference to any existing object
with the same type and value, while for mutable objects this is (strictly)?
not allowed.
</paraphrase>
Using the example given in 3.1, how do I verify that this is true pertaining
to immutable types? Below is my understanding on verifying the above statement:
>>> a = 1
>>> b = 1
>>> a is b
True
>>> id(a)
10901000
>>> id(b)
10901000
Is this correct?
Regards
Bernard
Dan Bishop
Yes, that's correct. However,
>>> a = 100
>>> b = 100
>>> a is b
False
>>> id(a)
135644988
>>> id(b)
135645000
Ben Finney
> <paraphrase>
> Depending on implementation, for immutable types, operations that
> compute new values may or may not actually return a reference to any
> existing object with the same type and value, while for mutable
> objects this is (strictly)? not allowed.
> </paraphrase>
>
> Using the example given in 3.1, how do I verify that this is true
> pertaining to immutable types?
You don't. By the language definition, it's entirely up to the
implementation whether *and when* to do this.
So, even if you find a particular session that does this, there's no
telling whether it'll stop happening at some future session using
*exactly the same inputs* -- and, if it did change, that would also be
entirely within the definition of the language.
If something in a language specification says "this set of conditions
leads to undefined behaviour", or "this aspect is implementation
defined", then *absolutely any behaviour* that fits the rest of the
definition is allowed, even if that results in non-determinism from
the programmer's perspective.
In short: Don't ever rely on such behaviour labelled with these "may
or may not" phrases, even if you run some tests and appear to get
predictable results.
--
\ “The power of accurate observation is frequently called |
`\ cynicism by those who don't have it.” —George Bernard Shaw |
_o__) |
Ben Finney
cokof...@gmail.com
> >>> b = 1
> >>> a is b
> True
> >>> id(a)
> 10901000
> >>> id(b)
> 10901000
Isn't this because integers up to a certain range are held in a single
memory location, thus why they are the same?
Duncan Booth
Yes, in *some* implementations of Python this is exactly what happens. The
exact range, or indeed whether it happens at all, and (if it does happen)
whether the range depends on the phase of the moon are all undefined.
This, for example, is what I just got for a similar sequence:
>>> a = 1
>>> b = 5-4
>>> a is b
False
>>> id(a)
43L
>>> id(b)
44L
>>>
Diez B. Roggisch
As the OP said:
<paraphrase>
Depending on implementation, for immutable types, operations that compute
new values may or may not actually return a reference to any existing object
with the same type and value, while for mutable objects this is (strictly)?
not allowed.
</paraphrase>
Which is exactly what happens - the actual implementation chose to cache
some values based on heuristics or common sense - but no guarantees are
made in either way.
Diez
Duncan Booth
> Which is exactly what happens - the actual implementation chose to cache
> some values based on heuristics or common sense - but no guarantees are
> made in either way.
Here's a puzzle for those who think they know Python:
Given that I masked out part of the input, which version(s) of Python might
give the following output, and what might I have replaced by asterisks?
>>> a = 1
>>> b = 5-4
>>> if a is b: print 'yes!'
...
yes!
>>> b = ****
>>> if a is b: print 'yes!'
...
>>> b
1
>>> type(b)
<type 'int'>
Steven D'Aprano
> Here's a puzzle for those who think they know Python:
>
> Given that I masked out part of the input, which version(s) of Python
> might give the following output, and what might I have replaced by
> asterisks?
There's too many variables -- at least five Python implementations that I
know of (CPython, Jython, PyPy, IronPython, and the Lisp-based
implementation that I can never remember the name of), and given that
this is an implementation-dependent feature it could have changed at any
time, in any version number (say, between minor releases). And there's
literally an infinite number of ways to get b equal to an int with the
value 1.
So I think unless somebody happens to have stumbled across this
behaviour, it's not predictable.
But having said that, I'm going to take a stab in the dark:
The line "b = ****" should be "b = int('1')"
and the version is CPython 1.4.
Am I close?
--
Steven
Stargaming
>
> Given that I masked out part of the input, which version(s) of Python
> might give the following output, and what might I have replaced by
> asterisks?
>
>>>> a = 1
>>>> if a is b: print 'yes!'
> ...
>>>> b
> 1
>>>> type(b)
> <type 'int'>
I know it! Using CPython 2.5.2a0, 2.6a1+ and 3.0a3+::
>>> b = type('type', (type,), {'__repr__': lambda self:
... "<type 'int'>"})('int', (object,), {'__repr__':
... lambda self:"1"})()
>>> b
1
>>> type(b)
<type 'int'>
>>> 1 is b
False
What's my prize?
Duncan Booth
> On Mon, 17 Mar 2008 10:40:43 +0000, Duncan Booth wrote:
>
>> Here's a puzzle for those who think they know Python:
>>
>> Given that I masked out part of the input, which version(s) of Python
>> might give the following output, and what might I have replaced by
>> asterisks?
>
> There's too many variables -- at least five Python implementations
> that I know of (CPython, Jython, PyPy, IronPython, and the Lisp-based
> implementation that I can never remember the name of), and given that
> this is an implementation-dependent feature it could have changed at
> any time, in any version number (say, between minor releases). And
> there's literally an infinite number of ways to get b equal to an int
> with the value 1.
True, there are a lot of variables, but perhaps not as many as you think.
For example, you can't get that output from IronPython or PyPy (or at least
not the versions I have kicking around) as they won't print 'yes!' for the
first test. You are correct though it is possible with both CPython and
Jython.
> So I think unless somebody happens to have stumbled across this
> behaviour, it's not predictable.
>
> But having said that, I'm going to take a stab in the dark:
>
> The line "b = ****" should be "b = int('1')"
>
> and the version is CPython 1.4.
>
> Am I close?
>
I don't have a copy of 1.4 to check so I'll believe you, but you can
certainly get the output I asked for with much more recent versions.
For the answer I actually want each asterisk substitutes for exactly one
character.
Duncan Booth
Ok, that one is impressive, I should have thought of putting some more
checks (like asserting type(a) is type(b)), but then the whole point of
setting the puzzle was that I expected it to be ripped to shreds. You
can queue up behind Steven to choose any of the exciting prizes on my
non-existent prize table.
Meanwhile, I'll fall back on the same quibbling getout I used with
Steven: too long.
Stargaming
> For the answer I actually want each asterisk substitutes for exactly one
> character.
Played around a bit and found that one:
Python 3.0a3+ (py3k:61352, Mar 12 2008, 12:58:20)
[GCC 4.2.3 20080114 (prerelease) (Debian 4.2.2-7)] on linux2
Type "help", "copyright", "credits" or "license" for more information.
>>> a = 1
>>> b = 1//1
>>> if a is b: print('yes!')
Duncan Booth
"the whole point of setting the puzzle was that I expected it to be ripped
to shreds", thanks for supporting my expectations.
Matthew Woodcraft
Duncan Booth <duncan...@suttoncourtenay.org.uk> wrote:
> I don't have a copy of 1.4 to check so I'll believe you, but you can
> certainly get the output I asked for with much more recent versions.
> For the answer I actually want each asterisk substitutes for exactly one
> character.
Then I'll guess you're looking for something like '0==0', with Python
2.2 or so (so that you'd get the PyTrue object).
-M-
Duncan Booth
Yes, that was the answer I was looking for: CPython 1.5.x (or maybe
earlier?) to 2.2.x without a bool type but separate values for true and
false (as used by PyWin). Also it seems that Jython 2.2.1 (and probably
other versions) behaves the same way.
I'm intrigued though by Stargaming's answer with 1//1, I must look into
that one.
Duncan Booth
I've had a look to see why this happens: long division (and in Python 3 all
integers are longs) allocates a new long to hold the result of the division
so it will never use one of the preallocated 'small int' values.
That makes sense so far as it goes, but I'm slightly suprised if it isn't
worth an extra check somewhere for numerator fitting in a machine int and
shortcutting the long division.