Hash merging behavior

Question

Is this behavior correct? I'm running some code like the following:

@a_hash = {:a => 1}
x = @a_hash
x.merge!({:b => 2})

At the end of all that, x's value has been changed as expected but so has the value for @a_hash. I'm getting {:a => 1, :b => 2} as the value for both of them. Is this normal behavior in Ruby?

score 4 · Accepted Answer · answered Feb 03 '12 at 19:46

Yes, instance variable @a_hash and local variable x store the reference to the same Hash instance and when you change this instance (using mutator method merge! that change object in place), these variables will be evaluated to the same value.

You may want to use merge method that create a copy of the object and don't change original one:

@a_hash = {:a => 1}
x = @a_hash
y = x.merge({:b => 2})
# y => {:a => 1, :b => 2}
# x and @a_hash => {:a => 1}

Ah, okay. I suspected that was what's happening. Thanks. – blim8183 Feb 03 '12 at 19:49 — blim8183, Feb 03 '12 at 19:49

score 3 · Answer 2 · answered Feb 03 '12 at 19:48

3

@a_hash is link to x. So if You want @a_hash not changed, you should do like this:

@a_hash = {:a => 1}
x = @a_hash.clone
x.merge!({:b => 2})

answered Feb 03 '12 at 19:48

alexkv

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2

I wouldn't recommend to clone when it's totally equivalent to `x = @a_hash.merge(:b => 2)` – tokland Feb 03 '12 at 20:22

score 1 · Answer 3 · answered Feb 03 '12 at 19:49

Yes, that's normal behavior in ruby (and most other languages). Both x and @a_hash are references to the same object. By calling merge! you change that object and that change in visible through all variables that refer to it.

If you don't want that behavior, you should either not use mutating methods (i.e. use x = x.merge(...) instead) or copy the object before you mutate it (i.e. x = @a_hash.dup).

Hash merging behavior

3 Answers3