[plt-scheme] keywords (a backward-incompatible change)

On Oct  1, Dave Herman wrote:
> Can we assume that:
> 
>      (symbol? (string->symbol ":a")) ; => #t

Yes, `string->symbol' always creates symbols.  Additional
`string->keyword' and back are used for keywords.


>      (symbol? '|:a|)                 ; => #t
>      (lambda (|:a|) |:a|)            ; => #<procedure>

Yes -- this is implied in this line from Matthew's message:

> (keyword? '|:a b|) ; => #f

The same should work (I think) for (lambda (\:a) \:a)

-- 
          ((lambda (x) (x x)) (lambda (x) (x x)))          Eli Barzilay:
                  http://www.barzilay.org/                 Maze is Life!