7 hours ago, Robby Findler wrote:
> On Sat, Nov 24, 2012 at 8:53 AM, Laurent <laurent.ors...@gmail.com> wrote:
> >
> > I think it would already be good enough to only consider the case
> > where the user types the paren and wants to remove them
> > immediately (e.g., he placed them in the wrong place, or wanted
> > square brackets instead, or just changed his mind). In the case
> > of meanwhile edits, I don't think the user would bother deleting
> > the closing paren himself.
>
> I think that hidden state like this can lead to confusing behavior.
> Probably you want to have deleting a paren do the same thing,
> regardless of what the character most recently typed was.
+1 for avoiding such hidden state. IIUC the xcode thing that was
mentioned tries to improve it by having a visible indication for the
stae, but I still think that the paredit behavr is better here -- when
you press backspace, this is what happens:
* If the preceding character is not a paren, it's removed as usual
* If it's a close paren, the cursor just goes to its left
* If it's an open parn then
- if the corresponding closing paren is on the right of the
cursor, delete both
- otherwise do nothing
The delete key works similarly in the other direction.
To demonstrate it, after I type "(define(foo x)(+ x 1", I get this:
(define (foo x) (+ x 1)|)
and hitting backspace now goes as follows:
(define (foo x) (+ x 1)|)
(define (foo x) (+ x 1|))
(define (foo x) (+ x |))
(define (foo x) (+ x|))
(define (foo x) (+ |))
(define (foo x) (+|))
(define (foo x) (|))
(define (foo x) |)
(define (foo x)|)
(define (foo x|))
(define (foo |))
(define (foo|))
(define (fo|))
(define (f|))
(define (|))
(define |)
(define|)
(defin|)
(defi|)
(def|)
(de|)
(d|)
(|)
; and finally nothing
--
((lambda (x) (x x)) (lambda (x) (x x))) Eli Barzilay:
http://barzilay.org/ Maze is Life!
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