IIRC, Ruby builds an s-expression tree under the hood (disclaimer: this is going by what I've read and I can't find the article right now, please take with seve

IIRC, Ruby builds an s-expression tree under the hood (disclaimer: this is going by what I've read and I can't find the article right now, please take with several grains of salt). I do know this, however: In Ruby, everything is an expression. So that's why code like.. foo = if bar? :bar else foo_proxy || :foo end ..can be expected to work. One thing I love about Ruby is that I feel as though it works "like a Lisp" under the hood, but doesn't look like one at all. The one thing I could never get over with Lisp-like languages was how difficult the syntax was for me to read. Ruby, OTOH, is a very familiar syntax and includes familiar concepts like various OOP encapsulation features, but it's all built around this expression-driven language that's both syntactically powerful and very easy to read. edit: I still wouldn't consider it part of the Lisp family, only heavily influenced by that kind of programming...