Hello Jow Forums, do you know Python?
If so, then why does the following code:
arr = [[0] * 2] * 3
[print (arr [x][0]) for x in range (3)]
arr [0][0] = 1
print ("New:")
[print (arr [x][0]) for x in range (3)]
print:
0
0
0
New:
1
1
1
?
Jow Forums
Camden Harris
Jacob Thomas
I think it has something to do with pointers
Gavin Cook
How to deal with pointer in python?
Jose Cox
You can't
Adam Garcia
Well what is it meant to print?
Mason Taylor
0 x 2 = 0 x 3 = 0
It'll print 0 three times since you have set the range to 3
You then set the array to 1 so it will print out 1 3 times?
Easton White
I don't set the array:
arr[0][0] = 1
sets only the first element of the first row and column
1
0
0
Hudson Richardson
Because you didn't make a deep copy. It would be inefficient to do that automatically.
Brandon Perry
Here user, im not sure what you're trying to do but you're just making it print out the value you have posted
Daniel Reyes
To complete my answer, look up the concept of reference counting. Instead of copying the content of the object itself, you just copy its address to save memory and cpu cycles. This way there will be several references pointing to the same memory area.