Array length with range operator

[gurpreetsingh13]

Hi Monks, I believe this must be something very simple..might be I am missing something.

For Array length, we use either $#<array-name> or scalar(<array>) or something with print to force a scalar context.

How can we create runtime arrays using range operator and get immediate length of the same.

Tried a few things and searched too, but can't find the solution.

This works-

perl -e 'my @a="aa".."zz"; print "Length=".@a;'

This also works-

perl -e 'my @a="aa".."zz"; print scalar(@a);'

This also works-

perl -e 'my @a="aa".."zz"; print @#a;'

Just if I need to do it in a single statement..

Syntax error thrown-

perl -e 'print length($("aa".."zz"));'

Syntax error thrown

perl -e 'print length(@("aa".."zz"));'

Even if I force scalar context,syntax error thrown.

perl -e 'my $len=@("aa".."zz");print $len;'

Might be I am forgetting something basic or so. Need your help please.

PS: Please don't advice on strict and warnings etc. This is just something quick and dirty.

[Happy-the-monk]

or try this ;-)

perl -wE'say scalar @{["aa".."zz"]}'

Cheers, Sören

[gurpreetsingh13]

Thanks.

Good trick.

Anonymous array - then convert the ref. back to array and apply scalar to it.

[AnomalousMonk]

Also works for "highest index" rather than "number of elements":

>perl -wMstrict -lE
"say scalar $#{[ 'aa' .. 'zz' ]};
"
675

[Athanasius]

16:35 >perl -wE "say scalar(my @a = 'aa' .. 'zz');"
676

16:35 >

Hope that helps,

[gurpreetsingh13]

Works. But I didn't want to generate a separate list variable @a.

[hdb]

perl -wE "say scalar( () = 'aa' .. 'zz');"

[kcott]

G'day gurpreetsingh13,

The same sort of question was asked last week: "Scalar context with list and range operator". The answers given there apply equally here.

[hbm]

perl -E"say@{[aa..zz]}+0"
676