From: "jeremyevans0 (Jeremy Evans) via ruby-core" Date: 2023-10-18T16:03:39+00:00 Subject: [ruby-core:115086] [Ruby master Bug#19932] Regexp o modifier freeze interpolation between method calls Issue #19932 has been updated by jeremyevans0 (Jeremy Evans). I don't think the documentation could be made significantly clearer, though potentially it could benefit from an additional example. It's expected that the `o` modifier works the way it does, and does not generate a new regexp with each call to the `poc` method in your example. There isn't a modifier that operates like `o`, but generates a new regexp for each call to `poc`, but not a new regexp per iteration inside `poc`. Doing so would be hard to reason about, because the `o` modifier changes how the code is compiled. What if the code was changed to?: ```ruby poc = lambda do |regexp| hs = [ 'azerty', 'azertyui', 'azertyuiop' ] out = [] hs.each do |h| out << h if /#{regexp}/o.match?('azerty') end out end ``` Should that still generate a new regexp per call to `poc`? Remember that `lambda` is not a keyword, it is a normal method call, so you can redefine it to use `define_method`, or to be the same as `Array#each`. For your example, for what you want, you should create a literal regexp without `o` modifier before the loop, set that to a local variable, and use that local variable inside the loop. ---------------------------------------- Bug #19932: Regexp o modifier freeze interpolation between method calls https://bugs.ruby-lang.org/issues/19932#change-104958 * Author: noraj-acceis (Alexandre ZANNI) * Status: Closed * Priority: Normal * ruby -v: ruby 3.2.2 (2023-03-30 revision e51014f9c0) [x86_64-linux] * Backport: 3.0: UNKNOWN, 3.1: UNKNOWN, 3.2: UNKNOWN ---------------------------------------- Taken the following PoC: ```ruby def poc(regexp) hs = [ 'azerty', 'azertyui', 'azertyuiop' ] out = [] hs.each do |h| out << h if /#{regexp}/.match?('azerty') end out end p poc('az') p poc('wxc') ``` I have the following output: ```ruby ["azerty", "azertyui", "azertyuiop"] [] ``` Now because the regexp never change inside the method once set I could add the `o` modifier so that `#{}` interpolation is computed only once. So the PoC becomes: ```ruby def poc(regexp) hs = [ 'azerty', 'azertyui', 'azertyuiop' ] out = [] hs.each do |h| out << h if /#{regexp}/o.match?('azerty') end out end p poc('az') p poc('wxc') ``` Output: ``` ["azerty", "azertyui", "azertyuiop"] ["azerty", "azertyui", "azertyuiop"] ``` So each future call to the method will be equals to the result of the first call because the regexp is "frozen". I was expecting the regexp to be "frozen" only for the current call in an effort of performance but instead it's "frozen" for future calls too. Another example of usage of `o` modifier here: https://learnbyexample.github.io/Ruby_Regexp/modifiers.html#o-modifier So is that a side effect or is that the expected behavior? -- https://bugs.ruby-lang.org/ ______________________________________________ ruby-core mailing list -- ruby-core@ml.ruby-lang.org To unsubscribe send an email to ruby-core-leave@ml.ruby-lang.org ruby-core info -- https://ml.ruby-lang.org/mailman3/postorius/lists/ruby-core.ml.ruby-lang.org/