PERL Miscellaneous - question about $list[1] x @list2

Author

question about $list[1] x @list2

jialin

2006-09-26, 9:59 pm

@list1 = qw/aa ab ac/;
@list2 = qw/1 2 3 4 5/;
print $list1[2] x @list2,"\n";

========
i know the output will be "acacacacac",
but what this x means in $list1[2] x @list2,

thanks,
Jialin

jialin

2006-09-26, 9:59 pm

I found the answer from perlop

Binary "x" is the repetition operator. In scalar context or if
the
left operand is not enclosed in parentheses, it returns a string
con-
sisting of the left operand repeated the number of times
specified by
the right operand. In list context, if the left operand is
enclosed in
parentheses, it repeats the list. If the right operand is zero
or neg-
ative, it returns an empty string or an empty list, depending on
the
context.

print '-' x 80; # print row of dashes

print "\t" x ($tab/8), ' ' x ($tab%8); # tab over

@ones = (1) x 80; # a list of 80 1's
@ones = (5) x @ones; # set all elements to 5

jialin wrote:
> @list1 = qw/aa ab ac/;
> @list2 = qw/1 2 3 4 5/;
> print $list1[2] x @list2,"\n";
>
> ========
> i know the output will be "acacacacac",
> but what this x means in $list1[2] x @list2,
>
> thanks,
> Jialin

Michele Dondi

2006-09-27, 4:00 am

On 26 Sep 2006 20:09:57 -0700, "jialin" <JialinLi1981@gmail.com>
wrote:

>i know the output will be "acacacacac",
>but what this x means in $list1[2] x @list2,

perldoc perlop

and search for 'Binary "x"'.

Michele
--
{$_=pack'B8'x25,unpack'A8'x32,$a^=sub{po
p^pop}->(map substr
(($a||=join'',map--$|x$_,(unpack'w',unpack'u','G^<R<Y]*YB='
..'KYU;*EVH[.FHF2W+#"\Z*5TI/ER<Z`S(G.DZZ9OX0Z')=~/./g)x2,$_,
256),7,249);s/[^\w,]/ /g;$ \=/^J/?$/:"\r";print,redo}#JAPH,

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