Re: [Readable-discuss] Proposal: allow closing SUBLIST by dedenting

2013-02-25 Thread David A. Wheeler
I said: Once it's removed from the BNF, I think we need to make sure that the Scheme specifically checks for it and forbids it. That way, people won't accidentally use such constructs for now, and that'll make it easier to add later if indeed it's added later. Alan Manuel Gloria: I

Re: [Readable-discuss] Proposal: allow closing SUBLIST by dedenting

2013-02-24 Thread David A. Wheeler
Beni Cherniavsky-Paskin: Here is yet another idea for opening multiple levels on one line, that does NOT involve column counting, only comparison of leading whitespaces. It's a backward-compatible extension to SUBLIST (similarly applicable to any competing FOOLIST semantics), so we could

Re: [Readable-discuss] Proposal: allow closing SUBLIST by dedenting

2013-02-24 Thread David A. Wheeler
Alan Manuel Gloria: Awww. (^.^)v :-). Since you're unwilling to put the full Beni formulation in *right now*, then let's use classic formulation. I hope somebody else speaks up for the full Beni formulation and gives a fully interesting example *real soon now* (^_^) Absolutely!

Re: [Readable-discuss] Proposal: allow closing SUBLIST by dedenting

2013-02-19 Thread Alan Manuel Gloria
On 2/19/13, Beni Cherniavsky-Paskin c...@users.sf.net wrote: Here is yet another idea for opening multiple levels on one line, that does NOT involve column counting, only comparison of leading whitespaces. It's a backward-compatible extension to SUBLIST (similarly applicable to any competing

Re: [Readable-discuss] Proposal: allow closing SUBLIST by dedenting

2013-02-19 Thread Alan Manuel Gloria
let !!$ x !!compute 'x y !!compute 'y !!use x y == let INDENT INDENT x ; stack: (0 2 ?) INDENT compute 'x ; stack: (0 2 ? 6) DEDENT ; stack: (0 2 ?), indentation = 4 y ; stack: (0 2 4) INDENT compute 'y ; stack (0 2 4 6) DEDENT DEDENT ; stack (0 2) use x y DEDENT == let

Re: [Readable-discuss] Proposal: allow closing SUBLIST by dedenting

2013-02-19 Thread Alan Manuel Gloria
Okay, okay, currently: let $ x $ compute 'x use x == (let ((x (compute 'x)) (use x)) So, let's try the new formulation: let !!$ x $ compute 'x !!use x == let INDENT INDENT x INDENT compute 'x ; stack: (0 2 ? ?) DEDENT DEDENT ; stack: (0 2), indentation = 2 use x DEDENT === let