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 2 years ago
Need help on 6.00 Handout, Lecture 5:
Here are the lists that I am looking at:
1.
L1 = [2]
L2 = [L1, L1]
print 'L2 =', L2
# L2 = [[2], [2]]
2.
L1[0] = 3
print 'L2 =', L2
# L2 = [[3], [3]]
3.
L2[0] = 'a'
print 'L2 =', L2
# L2 = ['a', [3]]
4.
L1 = [2]
L2 = L1
L2[0] = 'a'
print 'L1 =', L1
print 'L2 =', L2
# L1 = ['a']
# L2 = ['a']
I understand everything up to number 4 (correct answers start with a #). I can't understand why L1 = 'a' here. I was expecting L1 to be equal to [2].
Could someone write down an explanation about what is going on in part 4 of the
 2 years ago
Need help on 6.00 Handout, Lecture 5: Here are the lists that I am looking at: 1. L1 = [2] L2 = [L1, L1] print 'L2 =', L2 # L2 = [[2], [2]] 2. L1[0] = 3 print 'L2 =', L2 # L2 = [[3], [3]] 3. L2[0] = 'a' print 'L2 =', L2 # L2 = ['a', [3]] 4. L1 = [2] L2 = L1 L2[0] = 'a' print 'L1 =', L1 print 'L2 =', L2 # L1 = ['a'] # L2 = ['a'] I understand everything up to number 4 (correct answers start with a #). I can't understand why L1 = 'a' here. I was expecting L1 to be equal to [2]. Could someone write down an explanation about what is going on in part 4 of the

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JuanV
 2 years ago
Best ResponseYou've already chosen the best response.0L2 and L2 are pointing to the same object in memory  This is why around 30mins in the video the programming keeps running. This is call Aliasing  one object with more then one name.

RoamingBlue
 2 years ago
Best ResponseYou've already chosen the best response.0@tcc8m the first line sets L1 equal to a list that's big enough to contain two items. The second line sets L2 equal to L1 which means that they're both "equal to" or "pointing at" the same list. So when the third line sets the first member of the list to the value of 'a', both variables (since they're both pointing at /equal to the same list) get updated.

bwCA
 2 years ago
Best ResponseYou've already chosen the best response.1http://python.net/~goodger/projects/pycon/2007/idiomatic/handout.html#otherlanguageshavevariables

RoamingBlue
 2 years ago
Best ResponseYou've already chosen the best response.0@bwCA that's an awesome link. Thank you for that, I'm bookmarking it now :D

tcc8m
 2 years ago
Best ResponseYou've already chosen the best response.0Thanks @RoamingBlue, and @bwCA and @JuanV. Awesome answers! I agree with @RoamingBlue that the link was perfect. The price tags vs. box illustration is genius. Pictures were great too.

JuanV
 2 years ago
Best ResponseYou've already chosen the best response.0That is a great example bwCA :)
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