Trying to change single value in array created with iteration changes multiple values. Why?

Question

a = [[0, 0, 0], [0, 0, 0], [0, 0, 0]]
b = []
row = []
3.times { row << 0 }
3.times { b << row }

p a #=> [[0, 0, 0], [0, 0, 0], [0, 0, 0]]
p b #=> [[0, 0, 0], [0, 0, 0], [0, 0, 0]]
p a == b                         #=> true
p (a[1][1] = 1) == (b[1][1] = 1) #=> true

# and yet ...
a[1][1] = 1
b[1][1] = 1
p a #=> [[0, 0, 0], [0, 1, 0], [0, 0, 0]]
p b #=> [[0, 1, 0], [0, 1, 0], [0, 1, 0]]

With this code, I'd expect the modified b to be the same as the modified a, but instead, the second element of each nested array is modified.

What am I missing? Why do a[1][1] = 1 and b[1][1] = 1 produce different results?

All `row`s reference the same object. Use `3.times { b << row.dup }`. — Aleksei Matiushkin, Jun 12 '18 at 14:44

score 3 · Accepted Answer · edited Jun 12 '18 at 14:49

3

The line 3.times { b << row } adds 3 references to the same array to b, so modifying it once changes it three times in the array.

Try updating this to dup the row as you push it into b, i.e. 3.times { b << row.dup }.

(As I'm typing I'm seeing @mudasobwa beat me to it in a comment :) )

This approach will ensure b is filled with three unique arrays, rather than the same one three times.

If you're unsure, you can check the object_id, i.e. b.map(&:object_id) - at present, these will all be the same.

edited Jun 12 '18 at 14:49

mikej

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answered Jun 12 '18 at 14:44

SRack

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@SRack save for this question was asked here on SO a dozillion times and a lazy search would bring a dozens closed as duplicate answers. – Aleksei Matiushkin Jun 12 '18 at 15:25

Trying to change single value in array created with iteration changes multiple values. Why?

1 Answers1