CODEWITHSUNDEEP
perlif-statement
Daniel Camacho
Daniel Camacho

Reputation: 423

Evaluation if-statement in Perl

There are two declared variables, I want to evaluate at same time if both exist (are TRUE).

my @array = (1,2);
my $scalar = 0;

if (@array && $scalar){
  print "success\n";
}

&& does not work. Is there another operator or do I have to create 2 if-state?
I expected it to print success.

Upvotes: 0

Views: 375

Answers (3)

user2045980
user2045980

Reputation: 37

try this 

  my @array = (1,2);
  my $scalar = 0;

  if (scalar @array && defined $scalar){
     print "success\n";
  }

Upvotes: -1

Dave Sherohman
Dave Sherohman

Reputation: 46187

The code that does not work is not the code you posted.

perl -E 'my @array = (1,2); my $scalar = 1; if (@array && $scalar) { say "True!" } else { say "False" }'
True!

perl -E 'my @array = (); my $scalar = 1; if (@array && $scalar) { say "True!" } else { say "False" }'
False

Edit: More examples in response to kmxillo's comment:

perl -E 'my @array = (1,2); my $scalar = 0; if (@array && $scalar) { say "True!" } else { say "False" }'
False

perl -E 'my @array = (1,2); my $scalar = 0; if (@array && defined $scalar) { say "True!" } else { say "False" }'
True!

The number 0 is considered to be a false value (along with undef, the empty string, the empty array, and the string "0"), so setting $scalar to 0 makes @array && $scalar false. To keep it true when $scalar is 0, test defined $scalar instead.

Upvotes: 3

bjakubski
bjakubski

Reputation: 1747

If you consider array true when it has more than zero elements, then your if statement looks fine and should work as expected.

This program:

my @array = (1,2);
my $scalar = 1;

if (@array && $scalar) {
    print "TRUE!\n";
}

prints TRUE.

Why do you think && does not work?

Upvotes: 1

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