Hi,

In message "Re: Struggling with variable arguments to block"
    on 03/10/25, Austin Ziegler <austin / halostatue.ca> writes:

|> |You mean Hash#each and Hash::each_pair will behave
|> |differently after the fix??
|> Yes.
|
|What will Hash#each behave like?

The only notable change will be that hash.each{|*a| p a} will produce
[[k,v]], not [k,v].  Most code will run without any change.

							matz.