* Damian Conway ([EMAIL PROTECTED]) wrote:
I'm all hypered up. Does this mean that in Perl 6 I can do this:
@a =^+ 1;
You *can*, but it's not the same as doing:
@a ^=+ 1;
or:
@a ^+= 1;
%-)
What does this last line do? ;-)
--
Greg McCarroll
Damian Conway [EMAIL PROTECTED] writes:
@a =^+ 1; # evaluate each element of the list (1) in a numeric
# context and assign the resulting list to @a
# i.e. @a = $_ foreach +1;
@a ^=+ 1; # evaluate 1 in a numeric context
Quoting Leon Brocard ([EMAIL PROTECTED]):
Find out some more about Perl 6:
http://www.perl.com/pub/a/2001/10/02/apocalypse3.html
I'm all hypered up. Does this mean that in Perl 6 I can do this:
@a =^+ 1;
? I think I need a tranc now.
Cheers,
--
Merijn Broeren |
Software Geek | Yield to
Quoting Merijn Broeren ([EMAIL PROTECTED]):
Quoting Leon Brocard ([EMAIL PROTECTED]):
Find out some more about Perl 6:
http://www.perl.com/pub/a/2001/10/02/apocalypse3.html
I'm all hypered up. Does this mean that in Perl 6 I can do this:
@a =^+ 1;
? I think I need a tranc now.
