Re: Perl Array - What an array contains

[hippo]

Just test $#capitals

say $#capitals ? 'Lots of capitals' : "Just one: @capitals";

[pryrt]

Watch out, though. $#capitals would evaluate to true (-1) if the list has no capitals:

perl -lE "@capitals=(); $,=', '; print scalar(@capitals), $#capitals, 
+$#capitals ? 'Lots' : 'One';
0, -1, Lots

perl -lE "@capitals=qw(London); $,=', '; print scalar(@capitals), $#ca
+pitals, $#capitals ? 'Lots' : 'One';
1, 0, One

perl -lE "@capitals=qw(London Paris); $,=', '; print scalar(@capitals)
+, $#capitals, $#capitals ? 'Lots' : 'One';
2, 1, Lots

[AnomalousMonk]

There's yet another pitfall associated with the use of  $#array to determine the number of elements in an array. It's possible to assign to  $[ (see perlvar) to change the base index of Perl arrays. Doing so will result in your source entry appendage(s) receiving a sharp whack, so Don't Do That!^SM

Of course, it's possible to use  $#array for this purpose in a base-agnostic way:

c:\@Work\Perl\monks>perl -wMstrict -le
"print qq{perl array base is $[};
 ;;
 my @ra = ();
 print qq{A: array (@ra) is empty} if $#ra + 1 - $[ == 0;
 ;;
 @ra = (1);
 print qq{B: array (@ra) is empty} if $#ra + 1 - $[ == 0;
"
perl array base is 0
A: array () is empty

c:\@Work\Perl\monks>perl -wMstrict -le
"$[ = 1;
 print qq{perl array base is $[};
 ;;
 my @ra = ();
 print qq{A: array (@ra) is empty} if $#ra + 1 - $[ == 0;
 ;;
 @ra = (1);
 print qq{B: array (@ra) is empty} if $#ra + 1 - $[ == 0;
"
Use of assignment to $[ is deprecated at -e line 1.
perl array base is 1
A: array () is empty

But an array  @array evaluated in scalar context always yields the number of array elements, so I don't understand why one would ever do this in any other way.

BTW: It's possible to assign a number other than 0 or 1 to $[, but then you're really on thin ice, so Double-Plus-Don't Do That!

[hippo]

Indeed, but the possibility of "no capitals" was not in the spec. :-)