Question

When I try to pop an element from an array, it pops. When I assign that array to another variable before popping and then if I pop, the pop operation affects both the arrays.

For example:

 ruby-1.9.2-p290 :339 > a= [1,2,3]
     => [1, 2, 3] 
    ruby-1.9.2-p290 :340 > b = a
     => [1, 2, 3] 
    ruby-1.9.2-p290 :341 > a
     => [1, 2, 3] 
    ruby-1.9.2-p290 :342 > b
     => [1, 2, 3] 
    ruby-1.9.2-p290 :343 > a.pop
     => 3 
    ruby-1.9.2-p290 :344 > a
     => [1, 2] 
    ruby-1.9.2-p290 :345 > b
     => [1, 2] #WHY?
    ruby-1.9.2-p290 :346 > x = [1,2,3]
     => [1, 2, 3] 
    ruby-1.9.2-p290 :347 > y = x
     => [1, 2, 3] 
    ruby-1.9.2-p290 :348 > z = x
     => [1, 2, 3] 
    ruby-1.9.2-p290 :349 > y
     => [1, 2, 3] 
    ruby-1.9.2-p290 :350 > z
     => [1, 2, 3] 
    ruby-1.9.2-p290 :351 > y.pop
     => 3 
    ruby-1.9.2-p290 :352 > y
     => [1, 2] 
    ruby-1.9.2-p290 :353 > z
     => [1, 2] # WHY?
    ruby-1.9.2-p290 :354 > x
     => [1, 2] 
    ruby-1.9.2-p290 :355 > 

If I use pop, all the variables are affected. How do I keep the original array and pop from the other one only?

Answer 1

If you assign an array to a new variable, it’s not copied, but you only set a reference to the original array. If you want to keep the original array, you have to clone it by using dup:

ruby-1.9.2-p180 :001 > a = [1,2,3]
 => [1, 2, 3] 
ruby-1.9.2-p180 :002 > b = a.dup
 => [1, 2, 3] 
ruby-1.9.2-p180 :003 > b.pop
 => 3 
ruby-1.9.2-p180 :004 > a
 => [1, 2, 3] 
ruby-1.9.2-p180 :005 > b
 => [1, 2]