Perl - Array reference, using strict

Question

I have the following code:

my @array = ('a', 'b', 'c');

my $region = \@array;  # Returns an array reference
my $Value = ${@{$region}}[3];

I am using strict;

This code passed smoothly in Perl v5.8.6, and now that I installed v5.10.1, I get a runtime error:

Can't use string ("4") as an ARRAY ref while "strict refs" in use at ...

I changed the code to the following, and that solved the issue:

my @array = ('a', 'b', 'c');

my $region = \@Array;
my @List = @{$region};
my $Value = $List[3];

my question is, what's wrong with the previous way? What has changed between these two versions? What am I missing here?

Thanks, Gal

Also what I think the problem is... that your region contains 4 elements, than @{$region} is evaluated in a list context which returns 4 and your call becomes: $Value = ${'4'}[3] which produces the error — Dimitar Petrov
– Dimitar Petrov, Commented Aug 14, 2011 at 10:56
Why do you keep referencing the 4th element in a list which is three elements long in the code? — Conspicuous Compiler
– Conspicuous Compiler, Commented Aug 14, 2011 at 11:29

cjm · Accepted Answer · 2011-08-14 11:43:44Z

11

${@{$region}}[3] was never the correct way to access an arrayref. I'm not quite sure what it does mean, and I don't think Perl is either (hence the different behavior in different versions of Perl).

The correct ways are explained in perlref:

my $Value = ${$region}[3]; # This works with any expression returning an arrayref
my $Value = $$region[3];   # Since $region is a simple scalar variable,
                           # the braces are optional
my $Value = $region->[3];  # This is the way I would do it

answered Aug 14, 2011 at 11:43

cjm

62.2k9 gold badges133 silver badges179 bronze badges

Sign up to request clarification or add additional context in comments.

7 Comments

ysth Over a year ago

The new behavior is correct; the block after the $ should get scalar context. Older perl versions would erroneously promote the array to an array reference. But I thought this was going through a deprecation cycle - looks like that was done for @array->[] but not ${@array}[] :(

ikegami Over a year ago

${ @{$region} }[3] is ${ scalar(@{$region}) }[3]. It's equivalent to $0[3], $1[3], $2[3] or similar, depending on the number of elements in @$region.

ikegami Over a year ago

@Jared Ng, All 3? Only two ways were presented. If you consider adding parens, then there are far more than 3. ${$region}[3], ${region}->[3] and ($region)->[3], for starters.

ikegami Over a year ago

@ysth, Those curlies don't form a block.

ikegami Over a year ago

@ysth, as best I can tell, ${@array}[] didn't get a deprecation cycle cause it always worked correctly. (Worked in 5.10, don't have 5.8 anymore) @array->[] was deprecated in 5.12 and fixed in 5.14.

|

Jared Ng · Accepted Answer · 2011-08-14 10:51:36Z

2

This is how I would do it:

my @array = ('a', 'b', 'c');
my $region = \@array;
my $Value = $$region[1];
print $Value;

Output:

answered Aug 14, 2011 at 10:51

Jared Ng

5,0712 gold badges22 silver badges18 bronze badges

1 Comment

Jared Ng Over a year ago

Why exactly did this get a downvote? It's a perfectly valid method.

Collectives™ on Stack Overflow

Perl - Array reference, using strict

2 Answers 2

7 Comments

1 Comment

Your Answer

Hot Network Questions

Collectives™ on Stack Overflow

2 Answers 2

7 Comments

1 Comment

Your Answer

Sign up or log in

Post as a guest

Related