I am using the # frozen_string_literal: true magic comment that RuboCop enforces by default, and I can't append something to a string:
string = 'hello'
string << ' world'
because it errors out with:
can't modify frozen String (RuntimeError)
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1Why do you turn on frozen strings when you don't want your strings to be frozen?Jörg W Mittag– Jörg W Mittag2019-09-04 21:16:19 +00:00Commented Sep 4, 2019 at 21:16
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Because, as mentioned, it's enforced by default by RuboCop, and I tend to trust them, and because I do want some strings to be immutable.Iulian Onofrei– Iulian Onofrei2019-09-04 22:10:29 +00:00Commented Sep 4, 2019 at 22:10
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4 Answers
You add a + before the string like:
string = +'hello'
string << ' world'
puts(string)
hello world
1 Comment
cremno
Backwards compatible alternatives:
String.new('hello') or 'hello'.dupYou can also use +=:
s = 'H'
s += 'ello
=> "Hello"
Comments
frozen_string_literal is for string literal. To create a mutable String instance, use String#new or prepend a + to the string literal.
# frozen_string_literal: true
'foo'.frozen? # => true
String.new.frozen? # => false
(+'foo').frozen? # => false
2 Comments
Iulian Onofrei
What does this answer add different from this anwer and this comment?
Weihang Jian
1) I usually don't read all comments of all other answers, so do some people. 2) My answer is unique to other answers in this thread.
When # frozen_string_literal: true is enabled, you cannot 'mutate' any strings.
Want proof? With the following script...
# frozen_string_literal: true
str = 'hello'
str << ' world'
you will get the following error...
Traceback (most recent call last):
frozen_strings.rb:4:in `<main>': can't modify frozen String (FrozenError)
'Mutate' means change the value of the object.
Therefore, your example fails because << mutates strings and it is mutating your string string since you are using the << operator.
Want proof? Go into irb! Enter the follow:
str = 'hello' # => 'hello'
obj_id1 = str.object_id # => some number, ex: 12345
str << ' world' # => 'hello world'
obj_id2 = str.object_id # => some number, ex: 12345
obj_id1 == obj_id2 # this should return true, proving that you mutated the object
However, you can get around this by using str += ' world'.
Why? Because += is a shorthand reassignment. Instead of mutating, += creates a brand new string (with a brand new object_id) and stores it under the same variable name (in this case, str).
Want Proof? Check it out in irb!
str = 'hello' # => 'hello'
obj_id1 = str.object_id # => some number, ex: 12345
str += ' world' # => ' world'
obj_id2 = str.object_id # => some other number, ex: 356456345
obj_id1 == obj_id2 # => this returns false!
Let me know if this helped!
4 Comments
Iulian Onofrei
When
# frozen_string_literal: true is enabled, you can 'mutate' a string by using string << 'string' instead of string += 'string' that you mentioned.
Conturbo
This code gave me an error...
ruby # frozen_string_literal: true str = 'hello' str << ' world' ``` Traceback (most recent call last): frozen_strings.rb:4:in `<main>': can't modify frozen String (FrozenError) ```Conturbo
And besides, that example proves that you cannot mutate strings while
# frozen_string_literal: true is enabled, but in your first comment you're asserting that you can.Iulian Onofrei
I didn't want to come up as rude. But your last reply is still incorrect. Again, you can mutate strings, even when
frozen_string_literal is set to true, if you initialize it using = +, and not simply =.