Perl qq and underscore

Question

So I have two variables

$x = q(foo);
$y = q(bar);

My goal is to use them in a third variable with an underscore between them i.e. foo_bar. There are lots of ways to do this, but I wanted to use qq

so

$z = qq($x_$y);

This gives the following error

Global symbol "$x_" requires explicit package name at test.pl line 45.
Execution of C:\test.pl aborted due to compilation errors.

So I had to use curly brackets with the variable x to make it work

$z = qq(${x}_$y);

Why does underscore not work with qq? Why do I need curly brackets in this case?

`qq($x_$y)` is exactly the same as `$x_ . $y`. You need curlies because that's not what you want. — ikegami, Jun 23 '16 at 06:09
Re: `Why does underscore not work with qq?` It's not just `qq()`, the same issue you'll have with `"$x_$y";` for reason explained in answers. — mpapec, Jun 23 '16 at 07:29

score 7 · Accepted Answer · answered Jun 23 '16 at 05:35

That's because _ counts as a letter in identifiers (such as variable names).

When you write "$x_$y", Perl thinks you're trying to interpolate two variables, $x_ and $y. Similarly, when you write "$foo$bar", Perl thinks you're trying to interpolate $foo and $bar (not $f . 'oo' . $bar or $fo . 'o' . $bar or any other combination).

The general rule is: Names extend as far to the right as possible (that is, Perl chooses the longest possible interpretation for identifiers).

score 4 · Answer 2 · edited May 23 '17 at 10:28

4

The underscore _ is a legitimate character to use in variable names, so when you say "$x_$y" you are referring to (evaluating) a variable $x_, which clearly has not been defined.

See, for example, the accepted answer in this post for this and a lot more of this kind.

edited May 23 '17 at 10:28

Community

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answered Jun 23 '16 at 05:35

zdim

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Seriously, you are among the best of us here, and I will be pleased when your "score" matches your ability – Borodin Jun 23 '16 at 21:17
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@Borodin Oh, uh, that is way too kind of you to say, thank you. (Now _that_ is embarrassing.) Now I need to work really, really hard for it! So that perhaps some years down the line it's somewhat true :). Thank you for helping all along the way. I'll tidy up now, all of this is personal, so OT – zdim Jun 23 '16 at 21:21

score 3 · Answer 3 · answered Jun 23 '16 at 05:35

3

Because underscore may be part of the identifier. Otherwise, how parser should distinct $x from $x_ ?

So braces says that only x is identifer, not x_

answered Jun 23 '16 at 05:35

Alexandr Evstigneev

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Nagaraju · Answer 4 · 2016-06-23T05:44:42.557

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If you want it your way, then you should just escape the underscore character.

use strict;
use warnings;

my $x =qq(foo);
my $y=qq(bar);
my $z = qq($x\_$y);

print $z;

Ofcourse I would like to suggest that this is not the better way. The curly brace way is the most better approach.

edited Jun 23 '16 at 05:44

answered Jun 23 '16 at 05:37

Nagaraju

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score 1 · Answer 5 · answered Jun 23 '16 at 05:35

According to perl variable naming rull 'underscore' can be a part of the variable name. ref : http://sivasakthikumar.blogspot.in/2009/03/variable-names-can-start-with-letter_16.html

That's why its not able understand your variable name properly.So you need brackets.

Perl qq and underscore

5 Answers5