Default value of $#ArrayName when there is no element (Perl)

Question

The code below is to display the number of arguments entered in the command line.

#!/usr/bin/perl –w
$myVar = $#ARGV + 1;
print "Hi " , $ARGV[0] , "\n";
print "You have $myVar arguments\n";

From the perlintro, $#ARGV is a special variable which tells you the index of the last element of an array.

If this is the case, when I don't enter any value in the command line, how does $myVar value end up with 0 ?

Is it because when there is no element in the array, the index of "no element" is -1 ? As -1 + 1 = 0.

rjp · Accepted Answer · 2013-04-18 07:47:25Z

5

$#ARGV means "the index of the last element of ARGV" - not just any array as the perlintro sentence seems to imply.

For any array, if it's empty, $#array will be -1 and scalar @array will be 0.

CAVEAT: If someone has modified $[ ("Index of first element"), that'll change $# as well. You should probably always use scalar @array if you're after the length, and $array[-1] to get the last element.

> cat demo.pl
my @array = ();
print "Size=", scalar @array, " items, last=", $#array, "\n";
$[ = 2;
print "Size=", scalar @array, " items, last=", $#array, "\n";
> perl demo.pl
Size=0 items, last=-1
Size=0 items, last=1

answered Apr 18, 2013 at 7:47

rjp

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squiguy · Accepted Answer · 2013-04-18 07:37:42Z

2

According to the perlvar page:

@ARGV The array @ARGV contains the command-line arguments intended for the script. $#ARGV is generally the number of arguments minus one, because $ARGV[0] is the first argument, not the program's command name itself. See $0 for the command name.

answered Apr 18, 2013 at 7:37

squiguy

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Zaid · Accepted Answer · 2013-04-18 07:48:50Z

2

You are right.

$#ARGV is scalar @ARGV - 1, as squiguy points out.

But there are less-noisier alternatives to count the number of arguments passed to your program that you should consider using instead:

my $count = scalar @ARGV;  # Explicit using of 'scalar' function
my $count = 0+@ARGV;       # Implicitly enforce scalar context
my $count = @ARGV;         # Since the context is already set by LHS

answered Apr 18, 2013 at 7:48

Zaid

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2 Comments

rjp Over a year ago

$#ARGV won't always be scalar @ARGV-1 though. $# is an unreliable mistress and should be treated as if there's a sharp axe in play.

Zaid Over a year ago

@rjp : You're right that $#ARGV == @ARGV - 1 + $[;. The only thing stopping me from mentioning it is that $[ was deprecated in 5.12. As it stands, I would venture to say that 99.9% of the time $[ == 0, if not more

Matthias · Accepted Answer · 2013-04-18 07:49:08Z

0

Is it because when there is no element in the array, the index of "no element" is -1 ? As -1 + 1 = 0

Almost. It is not "the index of 'no element'" but rather the following rule applies:

perldata

The following is always true:

scalar(@whatever) == $#whatever + 1;

answered Apr 18, 2013 at 7:49

Matthias

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1 Comment

rjp Over a year ago

That won't be true if some (evil) person has changed $[.

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