Issue #21142 has been updated by Hanmac (Hans Mackowiak). You might want a different function. `Enumerable#each_with_index` does reset the Enumerator, while `Enumerator#with_index` does not. ```ruby p 3.times.lazy.take(0).each_with_index.to_a #[[0, 0], [1, 1], [2, 2]] p 3.times.lazy.take(0).with_index.to_a #[] ``` ---------------------------------------- Bug #21142: Lazy enumerator `.each_with_index` ignores `.take(0)` before it https://bugs.ruby-lang.org/issues/21142#change-112259 * Author: aaronkison (Aaron Kison) * Status: Open * ruby -v: 3.2.7 * Backport: 3.1: UNKNOWN, 3.2: UNKNOWN, 3.3: UNKNOWN, 3.4: UNKNOWN ---------------------------------------- Minimum code to produce problem: ``` class Numbers; def each; 100.times { yield _1 }; end; include Enumerable; end Numbers.new.lazy.take(0).each_with_index.map { _1 }.to_a ``` Output (at ruby 3.2.7, and 3.3.0): ``` [0, 1, ..., 99] ``` Expected output (and was as at ruby 3.1.4): ``` [] ``` It works when it opposite ordering: `Numbers.new.lazy.each_with_index.take(0).map { _1 }.to_a`. I suspect it may be related to the change here https://github.com/ruby/ruby/pull/11868/files but I'm not familiar with any of that code. It seems like it replaces an allocated index with a counting index, which my hunch is it works for every value except for 0. -- https://bugs.ruby-lang.org/