Question

I was fiddling around with id recently and realized that (c?)Python does something quite sensible: it ensures that small ints always have the same id.

>>> a, b, c, d, e = 1, 2, 3, 4, 5
>>> f, g, h, i, j = 1, 2, 3, 4, 5
>>> [id(x) == id(y) for x, y in zip([a, b, c, d, e], [f, g, h, i, j])]
[True, True, True, True, True]

But then it occurred to me to wonder whether the same is true for the results of mathematical operations. Turns out it is:

>>> nines = [(x + y, 9) for x, y in enumerate(reversed(range(10)))]
>>> [id(x) == id(y) for x, y in nines]
[True, True, True, True, True, True, True, True, True, True]

Seems like it starts failing at n=257…

>>> a, b = 200 + 56, 256
>>> id(a) == id(b)
True
>>> a, b = 200 + 57, 257
>>> id(a) == id(b)
False

But sometimes it still works even with larger numbers:

>>> [id(2 * x + y) == id(300 + x) for x, y in enumerate(reversed(range(301)))][:10]
[True, True, True, True, True, True, True, True, True, True]

What’s going on here? How does python do this?

Answer 1

Python keeps a pool of int objects in certain numbers. When you create one in that range, you actually get a reference to the pre-existing one. I suspect this is for optimization reasons.

For numbers outside the range of that pool, you appear to get back a new object whenever you try to make one.

$ python
Python 3.2 (r32:88445, Apr 15 2011, 11:09:05) 
[GCC 4.5.2 20110127 (prerelease)] on linux2
Type "help", "copyright", "credits" or "license" for more information.
>>> x = 300
>>> id(x)
140570345270544
>>> id(100+200)
140570372179568
>>> id(x*2)
140570345270512
>>> id(600)
140570345270576

Source

PyObject* PyInt_FromLong(long ival)
Return value: New reference. Create a
new integer object with a value of
ival.

The current implementation keeps an
array of integer objects for all
integers between -5 and 256, when you
create an int in that range you
actually just get back a reference to
the existing object. So it should be
possible to change the value of 1. I
suspect the behaviour of Python in
this case is undefined. 🙂

emphasis mine