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ViewsWhat happens in memory while unpacking a collection?
Say
list1=[4,8,12]
a,b,c=list1
output is a=4,b=8,c=12.
My confusion
Instructor told us that it is not like a gets mapped to 4, b to 8, and c to 12. I didn't understand what he told clearly (although I listened repeatedly to him multiple times). He was telling something like object is created for 4 and a is mapped to 4. But what is the difference between this and what I have presented below in figure?
2 answers
Answer question0
The thing about your picture that's misleading is that it implies that a, b, and c reference slices of list1. If you change list1, though, you will find that a, b, and c aren't affected by that change.
A better way to draw the picture might be to show 4, 8, and 12 separate from list1:
list1-->[ ][ ][ ]
| | |
V V V
4 8 12
^ ^ ^
| | |
a b c
All of the variables are independent of one another, even though some of them (e.g. list1[0] and a) currently point to the same values.
To put it another way: saying a = list1[0] is saying "evaluate list1[0] and assign a to reference whatever that value is right now", which is not the same as saying "make a be an alias for list1[0]".
0
Try this:
# define the list
list1=[4,8,12]
# reserve 3 memory spaces and unpack the values from list into them
# those memory spaces will contain one integer each one of the size of
# sys.getsizeof(int()) == 28 bytes (Python 3)
# a,b and c are actually pointers to those memory spaces
a,b,c=list1
print(a,b,c)
# change the first value of the list
list1[0] = 56
print(list1)
# now check that indeed "a" is not the same pointer than "list1[0]"
print(a)
But
You must to be careful with this kind of asignations with lists, try also this:
list2 = list1
print(list1, list2)
# then change any of them
list1 [0] = -1
# check that "list2" is pointing to the same memory address than "list1"
print(list1, list2)