Re: "<->" as "->" with automatic "is rw"

[Matt Diephouse]

On Fri, 20 Aug 2004 13:31:12 -0700, Larry Wall <[EMAIL PROTECTED]> wrote:
> It's vaguely possible I could be persuaded on the basis that
> 
>     for zip @a ¥ @b <-> { ($^a,$^b) = ($^b,$^a) }

Shouldn't that be:

    for zip @a, @b <-> { ... }
              --or--
    for @a ¥ @b <-> { ... }

?

--
matt