Perl: Creating new variables inside a loop

Question

I have code like the following:

 my $player_2ubid = grep { $_->ubid eq "2ubid" } @{$room_members };
 my $player_3ubid = grep { $_->ubid eq "3ubid" } @{$room_members };
 # ....
 my $player_11ubid = grep { $_->ubid eq "11ubid" } @{$room_members };

To avoid repetition, I want to go for a loop

for my $i ( 2 .. 11 ){
my $player_.$i.ubid = grep { $_->ubid eq "$i.ubid" } @{$room_members };
}

However this produces syntax error:

Can't modify concatenation (.) or string in scalar assignment at ***.t line 100, near "};"

What am I doing wrong here?

@KoVadim thanks. I am suspecting that $player_.$i.ubid may also be in trouble — ado
– ado, Commented Sep 4, 2013 at 7:20
Variable names with a number at the end is a really good indication that your code is designed badly. This should be an array or a hash. — Dave Cross
– Dave Cross, Commented Sep 4, 2013 at 10:40

ado · Accepted Answer · 2013-09-04 07:43:52Z

4

First you have to write $i."ubid" or "${i}ubid" instead of "$i.ubid".

And I'd use an hash like:

my %player_ubid;
for my $i ( 2 .. 4 ){
    $player_ubid{$i} = grep { $_->ubid eq $i."ubid" } @{$room_members };
}

edited Sep 4, 2013 at 7:43

ado

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answered Sep 4, 2013 at 7:33

Toto

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Comments

akashdeep · Accepted Answer · 2013-09-04 07:56:29Z

1

You are declaring the variable in wrong way, you can use hash instead of it.

But if you want to achieve in similar fashion, do it in this way

for my $i ( 2 .. 11 ){
  my $var = "player_".$i."ubid";
  $$var = grep { $_->ubid eq $i."ubid" } @{$room_members };
}

$$ creates the variable of the value inside the variable, just like in php.

edited Sep 4, 2013 at 7:56

answered Sep 4, 2013 at 7:31

akashdeep

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9 Comments

varnie Over a year ago

$$var will produce warnings when use strict, right? And, it is almost always a bad idea. But I understand you provided it as another way to solve the problem.

Samveen Over a year ago

What will be the scope of the variables created this way?

akashdeep Over a year ago

If you use strict then it is not a good approach, because scope of those variable remain inside the loop and also strict will produce warings when we use $$.

amon Over a year ago

To make this clear: strict has nothing against the $$var – but it forbids using the names of variables as references. Using names as references is a bad programming practice (see varnie's link), and (in Perl) can only create global variables, which have infinite scope.

TLP Over a year ago

@amon It is not true that only global variables can be created. Try my $var = 'Foo::foo';

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