From: Brian Ross Date: 2008-08-30T06:14:29+09:00 Subject: Re: Writing a method to handle a code block? ------=_Part_20598_28250285.1220044758956 Content-Type: text/plain; charset=ISO-8859-1 Content-Transfer-Encoding: 7bit Content-Disposition: inline Thank you, everyone. If I understand correctly: &foo just means that whatever is passed to it (& requires a code block) will be saved as a local variable that may be used later. The call method executes the block that it acts on and then executes it but first assigns the block's | block argument | to the parameter passed to it. Is it then that yield functions the same as the call method? I haven't quite gotten to proc objects yet, but it looks like using & converts the code block into an object, but yield, strictly, doesn't? Are code blocks the exception to the "everything in ruby is an object" rule? I'm certainly understanding this far better than I was previously, I think. Brian ------=_Part_20598_28250285.1220044758956--