The index expression of $array[ EXPR ] is evaluated in scalar context, and the an array in scalar context evaluates to the number of elements in the array.
my @folder = qw( a b c d e f );
my @name = qw( x y z );
my $var = $folder[@name]; # $var = 'd', since @name == 3.
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Thanks ikegami, for the wonderful explanation ! And I really appreciate the example. It really helps a lot.
And thanks for the other suggestions as well. I'm really glad I chanced upon this forum. I did thought of making use of print to try figuring out what the statements would do, but the Internet connection here is really awful. So I gave up downloading Strawberry Perl.
Only could resort to trying to understand the code using Notepad++.
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ikegami's post is technically correct.
However in the code that you presented, @name has nothing to do with @folder - this should be a "red flag"!
Something is missing here!
This uses the number of items in @name as an index into the @folder array.
While this could happen, it would be unusual absent other context.
I would be looking for some code that builds @folder based upon the elements in @name.
Update:
Furthermore, $var = $folder[@name]; would be unusual since Perl indicies are "zero-based".
$var = $folder[@name-1]; would I think, be more expected - another "red-flag".
You may sometimes see $#name instead of @name-1, but this means the same thing - the last index.
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suppose you have @a= qw(1 2 3 4);
you are writting print "\n @a"; this will print 1 2 3 4.
if you are trying to write like...
$p= @a, then it will give you count of array i.e. 4
so it does not mean that if we write @a as index inside array would give the value of 4th index of array. Basically if we write code..
$b[@a] with return nothing because bare @a return complete set of value which is not a valid index and that is passing as index in @b so of course will return blank.
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Perl syntax can be tricky. Does this help? Or just further confuse?
Update: I guess it confused one Monk, because I got a down vote.
I am genuinely trying to be helpful here.
When you write: $y[@x], you are accessing a single element of @y with the index of the scalar value of @x.
Note: using $a and $b or @a or @b is in general not a good idea as the $a and $b variables have special meaning to Perl and are used in sort. Better is to use: x,y,z.
However: "@a return complete set of value" this is not correct. What @a would mean is context specific.
#!/usr/bin/perl -w
use strict;
use Data::Dumper;
my @x= qw(1 2 3 4);
my @y= qw(7 8 9 10 11 12 13 14);
my $num = $y[@x]; # @x evaluates here to the number "4"!!!
# the number of elements in @x
# As said before this is a VERY unusual
# formulation!!!
# It looks weird because it is weird.
# As mentioned above more normal would be
# $y[@x-1] or $y[$#x]
# I'm not saying that $y[@x] cannot have a defined,
# value but just that it is "strange" absent other
# context and would be worthy of a comment in the
# code.
print "The number in y is: $num\n"; #the 5th thing in @y!! WHOA!
#print $x[@y]; # ERROR: Use of uninitialized value in print
# this is because: $x[8] is undefined!
# However consider this:
# here [@x] means something completely different
push @y, [@x]; # this means: assign new memory,
# copy @x into that new memory ,
# push a reference to that memory
# onto @y
print Dumper \@y;
__END__
The number in y is: 11 ### the 5th thing in @y
$VAR1 = [
'7',
'8',
'9',
'10',
'11',
'12',
'13',
'14',
[
'1',
'2',
'3',
'4'
]
];
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Damien have a look at Arrays: A Tutorial/Reference. The relevant section is "Get count of elements" but it is well worth reading through.
Often you can find your answer by adding a print statement, taking ikegami's example:
print "@name";
# prints x y z
print @name;
# prints 3
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print evaluates its operands in list context, so print @name; does not print 3. (print scalar(@name); and print 0+@name; do, though)
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Aagh - I did try it before I replied but like this:
print @name . "\n";
then I took the "\n" out to make it clearer (but didn't run it!). I take it the concatenation makes print evaluate in scalar context. So should be:
print "@name";
# prints x y z
print @name;
# prints xyz
print @name . "\n";
# prints 3
(lidden: I was still fixing this when you replied)
Update - then I found this:
If you believe in Lists in Scalar Context, Clap your Hands
I'll be back in a few weeks.
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use Data::Dumper;
....
print Dumper \@name;
Another module that you have to install before being able to use it:
use Data::Dump qw(pp);
.....
print pp \@name;
Basically, both of these modules can print/dump an arbitrary Perl data structure.
The format of the output is a bit different, but they both can do it.
This is actually pretty much magic.
It is of huge help when debugging.
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