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

开发者 https://www.devze.com 2023-03-29 09:14 出处:网络
Why is a variable name followed by an underscore not evaluated correctly during string interpolation in Perl?

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

my $i = 3;

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


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.


$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.


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


Always use these:

use strict;
use warnings;
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