How explicitly resolve variables in a perl string? - string

Question

In my perl script I want to have both versions of $config directory:
my $config='$home/client/config';
and
my $config_resolved="$home/client/config";
But I want to get $config_resolved from $config, i.e. something like this:
my $config_resolved=resolve_vars($config);
How can I do such thing in perl?

Answer 1

From the Perl FAQ (which every Perl programmer should read at least once):
How can I expand variables in text strings?
(contributed by brian d foy)
If you can avoid it, don't, or if you can
use a templating system, such as Text::Template or Template Toolkit,
do that instead. You might even be able to get the job done with
sprintf or printf:
my $string = sprintf 'Say hello to %s and %s', $foo, $bar;
However, for the one-off simple case where I don't want to pull out a
full templating system, I'll use a string that has two Perl scalar
variables in it. In this example, I want to expand $foo and $bar to
their variable's values:
my $foo = 'Fred';
my $bar = 'Barney';
$string = 'Say hello to $foo and $bar';
One way I can do this involves the substitution operator and a double /e flag. The
first /e evaluates $1 on the replacement side and turns it into $foo. The
second /e starts with $foo and replaces it with its value. $foo,
then, turns into 'Fred', and that's finally what's left in the string:
$string =~ s/(\$\w+)/$1/eeg; # 'Say hello to Fred and Barney'
The /e will also silently ignore violations of strict, replacing undefined
variable names with the empty string. Since I'm using the /e flag
(twice even!), I have all of the same security problems I have with
eval in its string form. If there's something odd in $foo, perhaps
something like #{[ system "rm -rf /" ]}, then I could get myself in
trouble.
To get around the security problem, I could also pull the
values from a hash instead of evaluating variable names. Using a
single /e, I can check the hash to ensure the value exists, and if it
doesn't, I can replace the missing value with a marker, in this case
??? to signal that I missed something:
my $string = 'This has $foo and $bar';
my %Replacements = (
foo => 'Fred',
);
# $string =~ s/\$(\w+)/$Replacements{$1}/g;
$string =~ s/\$(\w+)/
exists $Replacements{$1} ? $Replacements{$1} : '???'
/eg;
print $string;

Answer 2

I use eval for this.
So, you must replace all scalars (their names) with their values.
$config = 'stringone';
$boo = '$config/any/string';
$boo =~ s/(\$\w+)/eval($1)/eg;
print $boo;

Answer 3

Because you are using my to declare it as private variable, you might as well use a /ee modifier. This can find variables declared to be in local scope:
$boo =~ s/(\$\w+)/$1/eeg;

Answer 4

This is most tidily and safely done by the double-eval modifier on s///.
In the program below, the first /e evaluates the string $1 to get $home, while the second evaluates $home to get the variable's value HOME.
use strict;
my $home = 'HOME';
my $config = '$home/client/config';
my $config_resolved = resolve_vars($config);
print $config_resolved, "\n";
sub resolve_vars {
(my $str = shift) =~ s/(\$\w+)/$1/eeg;
return $str;
}
output
HOME/client/config