Question

I have just started playing with Ruby and I’m stuck on something. Is
there some trick to modify the casefold attribute of a Regexp object after
it’s been instantiated?

The best idea what I tried is the following:

irb(main):001:0> a = Regexp.new('a')
=> /a/
irb(main):002:0> aA = Regexp.new(a.to_s, Regexp::IGNORECASE)
=> /(?-mix:a)/i

But none of the below seems to work:

irb(main):003:0> a =~ 'a'
=> 0
irb(main):004:0> a =~ 'A'
=> nil
irb(main):005:0> aA =~ 'a'
=> 0
irb(main):006:0> aA =~ 'A'
=> nil

Something I don’t understand is happening here. Where did the ‘i’ go on line
8?

irb(main):07:0> aA = Regexp.new(a.to_s, Regexp::IGNORECASE)
=> /(?-mix:a)/i
irb(main):08:0> aA.to_s
=> "(?-mix:a)"
irb(main):09:0>

I am using Ruby 1.9.3.

I am also unable understand the below code: why returning false:

/(?i:a)/.casefold?      #=> false

Answer 1

As Frederick already explains, calling to_s on a regex will add modifiers around it that ensure that its properties like case-sensitiveness are preserved. So if you insert a case-sensitive regex into a case-insensitive regex, the inserted part will still be case-sensitive. Likewise the modifiers given to Regexp.new will have no effect if the first argument is a regex or the result of calling to_s on one.

To solve this issue, call source on the regex instead of to_s. Unlike to_s, source simply returns the source of regex without adding anything:

aA = Regexp.new(a.source, Regexp::IGNORECASE)

I am also unable understand the below code: why returning false:

/(?i:a)/.casefold? 

Because (?i:...) sets the i flag locally, not globally. It only applies to the part of the regex within the parentheses, not the whole regex. Of course in this case the whole regex is within the parentheses, but that doesn’t matter as far as methods like casefold? are concerned.