Why is a variable name followed by an underscore not evaluated correctly during string interpolation in Perl?

Question

my $i = 3;

print "i = $i\n"; # works, prints "i = 3"
print "_i = _$i\n"; # works, prints "_i = _3"
print "i_ = $i_\n"; # FAILS, prints "i_ = "
print "_i_ = _$i_\n"; # sort of works, prints "_i_ = _"

[Always use strict and always use warnings!](http://joelslinux.blogspot.com/2011/06/use-strict-and-warnings.html) — Joel Berger, Aug 21 '11 at 11:15

Alan Haggai Alavi · Answer 1 · 2013-04-05T01:08:06.810

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In addition to the other answers, you can use the alternative syntax for specifying variables:

print "i_ = ${i}_\n";

Note the usage of curly brackets: { and } to specify the variable name. Whenever in doubt, you may opt for this syntax.

edited Apr 05 '13 at 01:08

answered Aug 21 '11 at 11:18

Alan Haggai Alavi

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Mat · Answer 2 · 2011-08-21T11:13:10.760

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$i_ is a valid identifier, so it's trying to print the value of that variable (which you haven't set, so it is undef).

Turn on strict and warnings.

edited Aug 21 '11 at 11:13

answered Aug 21 '11 at 11:06

Mat

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I noticed your buddhabrot, Mat. Perhaps this would interest you: https://github.com/buddhabrot/buddhabrot (I couldn't reach you any other way so I had to find an innocent comment somewhere). – buddhabrot Dec 07 '11 at 16:48

score 7 · Answer 3 · answered Aug 21 '11 at 11:11

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Mat is right. If you really need that underscore immediately after the value use backslash: "$i\_".

answered Aug 21 '11 at 11:11

Itamar

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`print 'i_ = ' . $i . "_\n";` – Joel Berger Aug 21 '11 at 11:13
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I much prefer `${i}` over the backslash, since they do different things in other contexts. – tchrist Aug 21 '11 at 14:22
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A backslash suppressed interpolation. Curlies around a variable name interpolate the value of that variable. `print "${i}_\n"` is easier on my eyes and brain, because those two things would do the same thing if unquoted. – tchrist Aug 22 '11 at 15:21

shawnhcorey · Answer 4 · 2011-08-21T13:35:18.707

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Always use these:

use strict;
use warnings;

edited Aug 21 '11 at 13:35

answered Aug 21 '11 at 12:31

shawnhcorey

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Why is a variable name followed by an underscore not evaluated correctly during string interpolation in Perl?

4 Answers4