Ruby freeze method

Ruby Freeze method done following things on different objects, basically it's make object constant or immutable in ruby.

1. String

str = "this is string"
str.freeze

str.replace("this is new string") #=> FrozenError (can't modify frozen String)
or
str[0] #=> 't'
str[0] = 'X' #=> FrozenError (can't modify frozen String)

But you can assign new string or change its reference. Once the reference change then it is not freeze (or constant) object.

str = "this is string"
str.freeze
str.frozen? #=> true
str = "this is new string"
str.frozen? #=> false

2. Array :-

arr = ['a', 'b', 'c']
arr.freeze

arr << 'd' #=> FrozenError (can't modify frozen Array)
arr[0] = 'd' #=> FrozenError (can't modify frozen Array)
arr[1].replace('e') #=> ["a", "e", "c"]
arr.frozen? #=> true

arr array is frozen means you can’t change the array. But the strings inside the array aren’t frozen. If you do a replace operation on the string “one”, mischievously turning it into “e”, the new contents of the string are revealed when you reexamine the (still frozen!) array.

3. Hash

hash = {a: '1', b: '2'}
hash.freeze

hash[:c] = '3' #=> FrozenError (can't modify frozen Hash)
hash.replace({c: '3'}) #=> FrozenError (can't modify frozen Hash)
hash.merge({c: '3'}) #=> return new hash {:a=>"1", :b=>"2", :c=>"3"}
hash.merge!({c: '4'}) #=> FrozenError (can't modify frozen Hash)
hash.frozen? #=> true

hash = {:a=>"1", :b=>"2", :c=>"3"}
hash.frozen? #=> false

So we can't modify the hash content but we can refer it to new hash (just like array)

• freeze - prevents modification to the Hash (returns the frozen object)
• [] - accesses a value from the hash
• stat.class.name.underscore.to_sym - I assume this returns a lowercase, snake case version of the given object's class name (underscore is not in the standard library, so I'm not completely sure)
• call invokes the lambda associated with stat.class.name.underscore.to_sym key.

For instance, passing ['foo', 'bar'] as the argument to track_for would invoke the send(stat[0], stat[1]) lambda.