[SOLVED] python reverse function

sharky · 06-25-2014, 12:03 PM

Is this the way reverse is supposed to work in python?

Quote:

> python
Python 2.7.3 (default, Feb 27 2014, 19:37:34)
[GCC 4.7.2] on linux2
Type "help", "copyright", "credits" or "license" for more information.
>>> mylist = [1,2,3]
>>> mylist
[1, 2, 3]
>>> copylist = mylist
>>> copylist
[1, 2, 3]
>>> copylist.reverse()
>>> mylist
[3, 2, 1]
>>>

Why does reversing the copy cause the original to get reversed?

dugan · 06-25-2014, 12:26 PM

It's not a matter of how reversing works. It's a matter of how assignments work.

Code:

# a refers to the location in memory where the list is 
a = [1, 2, 3]

# b refers to the same memory location as a
b = a

# the list at the memory location referred to by a (and therefore b) is reversed
a.reverse()

If you want b to be a copy of a, you need to specify that explicity. It's a list of integers, so we can use what's called a "shallow copy":

Code:

import copy
a = [1, 2, 3]
b = copy.copy(a)
b.reverse()

Note: this is actually fairly consistent with how it works in most other languages. "Complex types", such as ArrayLists, work like this in C# and Java, and you'd usually use pointers and/or references in C/C++ to get equivalent behavior. You get better performance that way.

sharky · 06-25-2014, 01:05 PM

Yep. You're right.

Verified it with another function.

Quote:

> python
Python 2.7.3 (default, Feb 27 2014, 19:37:34)
[GCC 4.7.2] on linux2
Type "help", "copyright", "credits" or "license" for more information.
>>> mylist = [1,2,3]
>>> copylist = mylist
>>> copylist.append(4)
>>> mylist
[1, 2, 3, 4]

I'm not sure I agree it should work this way but I understand what's going on now.

Thanks for the tip on shallow copy.