CODEWITHSUNDEEP
perlsyntax
cellis
cellis

Reputation: 79

Not understanding what the % is doing in Perl

I am reviewing a script and I have come across a block that I can't figure out:

if ((@ifAliasVals[1])&&(%{@ifAliasVals[0]}))  {

...

}

What is the %{@ifAliasVals[0]} doing?

Upvotes: 1

Views: 73

Answers (2)

Joe Z
Joe Z

Reputation: 17936

The syntax %{ ... } treats the expression between the braces as a hash-reference, and dereferences it, giving you the entire hash.

The syntax @ifAliasVals[0] returns an array slice from the array @ifAliasVals. In this case, the array slice only has one element, and so that would have been better written as $ifAliasVals[0]. In any case, it gives you the first element of the array @ifAliasVals.

The entire expression %{ @ifAliasVals[0] } therefore interprets the first element of @ifAliasVals as a hash reference, dereferencing it, yielding the contents of that hash.

In the context of your if-statement above, the right-hand of the && will be true if the hash has any elements, and will be false if the hash is empty.

Upvotes: 2

ikegami
ikegami

Reputation: 385847

if (@ifAliasVals[1] && %{ @ifAliasVals[0] })

is a bad way of writing

if ($ifAliasVals[1] && %{ $ifAliasVals[0] })

%{ EXPR } is a hash dereference. %{ EXPR } is basically the same as %hash, but accesses the via a reference rather than its name.

In scalar context, %hash returns true iff the hash has any elements, so %{ $ifAliasVals[0] } checks if the hash referenced by $ifAliasVals[0] has elements.

my $hash1_ref = { };
my $hash2_ref = { a => 1 };
say %{ $hash1_ref } ? 'has elements' : 'empty';  # empty
say %{ $hash2_ref } ? 'has elements' : 'empty';  # has elements

Upvotes: 4

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