Re^2: Problems with grep

Grep returns scalars, example

D:\>perl -e"die grep /6/, 3 .. 6,6,6"
666 at -e line 1.
[download]

grep returns a list, and in scalar context, the number of elements the expression was true.

In your example, die outputs the list returned by grep:

perl -e "die scalar grep /6/, 3 .. 6,6,6"
3 at -e line 1.
[download]

What you did is essentially the same as:

perl -e 'die qw/1 2 3/'
123 at -e line 1.
[download]

citromatik

Comment on Re^2: Problems with grep Select or Download Code

Replies are listed 'Best First'.
Re^3: Problems with grep by Anonymous Monk on Jan 30, 2009 at 14:58 UTC
`\| \\|/ scalars<-- /\|\ \|` [download] `perldoc -f grep` is better :)	[reply] [d/l] [select]
Re^4: Problems with grep by citromatik (Curate) on Jan 30, 2009 at 15:11 UTC
~~That is not correct, as I said before grep returns a list, and that list is not restricted to be a list of scalars as you are claiming:~~ update: References are scalars, sorry for the confusion `use strict; use warnings; use Data::Dumper; my @t = ( [1,2,3], 0, [4,5,6] ); my @k = grep { ref $_ eq "ARRAY" } @t; print Dumper \@k;` [download] Outputs: `531(255)[1603][/home/pignatelli/tmp]$ perl kkk.pl $VAR1 = [ [ 1, 2, 3 ], [ 4, 5, 6 ] ];` [download] citromatik	[reply] [d/l] [select]
Re^5: Problems with grep by ikegami (Patriarch) on Jan 30, 2009 at 15:31 UTC
that list is not restricted to be a list of scalars Yes it is, as all lists are. You can't have a list of anything but scalars. In your example, grep returned two scalars, both of which are references.	[reply]