I'm having a little trouble comprehending this simple use of the /e regex modifier.
my $var = 'testing';
$_ = 'In this string we are $var the "e" modifier.';
s/(\$\w+)/$1/ee;
print;
Returns: "In this string we are testing the "e" modifier."
I cannot see why two 'e' modifiers are required. As far as I can see, $1 should capture '$var' from the string and a single 'e' modifier should then be able to replace the variable with its value. I must be misunderstanding something however, since trying the above code with just one 'e' modifier does not visibly replace anything in the string.
Excuse me for asking such a simple question!
Thanks.
It’s not exactly a “simple” question, so don’t beat yourself up.
The issue is that with a single /e
, the RHS is understood to be code whose eval
’d result is used for the replacement.
What is that RHS? It’s $1
. If you evaluated $1
, you find that contains the string $var
. It does not contain the contents of said variable, just $
followed by a v
followed by an a
followed by an r
.
Therefore you must evaluate it twice, once to turn $1
into $var
, then again to turn the previous result of $var
into the string "testing"
. You do that by having the double ee
modifier on the s
operator.
You can check this pretty easily by running it with one /e
versus with two of them. Here’s a demo a both, plus a third way that uses symbolic dereferencing — which, because it references the package symbol table, works on package variables only.
use v5.10;
our $str = q(In this string we are $var the "e" modifier.);
our $var = q(testing);
V1: {
local $_ = $str;
s/(\$\w+)/$1/e;
say "version 1: ", $_;
}
V2: {
local $_ = $str;
s/(\$\w+)/$1/ee;
say "version 2: ", $_;
}
V3: {
no strict "refs";
local $_ = $str;
s/\$(\w+)/$$1/e;
say "version 3: ", $_;
}
When run, that produces:
version 1: In this string we are $var the "e" modifier.
version 2: In this string we are testing the "e" modifier.
version 3: In this string we are testing the "e" modifier.