python slice method does not always return a new address in memory? [duplicate]

Question

x = [1,2]
for i in range(4):
    y = x[:]
    print id(y)

the results are like:

4392626008 4392835408 4392626008 4392835408

My purpose is to copy x each time and do something with a new container. My python is version 2.7.5 and OS system is Mac OS X 10.9, does it matter?

It could be that because the scope of y is only inside the for loop, and hence the memory address can be reused. — Kaarel Nummert
– Kaarel Nummert, Commented May 21, 2014 at 8:28
Why should y have a different address each time ??? What you are trying to do is unclear. — user1196549
– user1196549, Commented May 21, 2014 at 8:29
@AndyLiu you are getting a new container... It just happens to be re-using the address of an old one that you've informed Python it can garbage collect by not having any references to the previous created slice(s) any more — Jon Clements
– Jon Clements, Commented May 21, 2014 at 8:34

Jon Clements · Accepted Answer · 2014-05-21 08:28:26Z

4

You're never keeping all occurrences of y around - you're just rebinding the name y to a copy of x each time in the loop, so that by a later point in the loop - Python might well choose to reallocate the same area of memory for the new slice. And since id in CPython returns the memory address, you may get the same...

for i in range(4):
    # Rebinding `y` inside the loop - making the object available for garbage collection 
    y = x[:] 
    print id(y)

If you were to keep all y about, then you will get unique ids in CPython:

>>> x = [1, 2]
>>> ys = [x[:] for _ in range(4)]
>>> map(id, ys)
[40286328, 40287568, 40287688, 40287848]

answered May 21, 2014 at 8:28

Jon Clements

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