Question

I am studying about regular expression and struck with the

1. lookaround concept

   and

2. with few syntax.

After doing googling, I thought it is a right forum to ask for help.
Please help with this concept.

As I am not good with understanding the explanation.
It will be great if I get plenty of different examples to understand.

For me the modifer /e and || are new in regex please help me in understanding
the real use. Below is my Perl Script.

$INPUT1="WHAT TO SAY";
$INPUT2="SAY HI";
$INPUT3="NOW SAY![BYE]";
$INPUT4="SAYO NARA![BYE]";

$INPUT1=~s/SAY/"XYZ"/e;   # /e What is this modifier is for

$INPUT2=~s/HI/"XYZ"/;

$INPUT3=~s/(?<=\[)(\w+)(?=])/ "123"|| $1 /e; #What is '||' is use for and what its name
$INPUT4=~s/BYE/"123"/e;

print "\n\nINPUT1 = $INPUT1 \n \n ";
print "\n\nINPUT2 = $INPUT2 \n \n ";
print "\n\nINPUT3 = $INPUT3 \n \n ";
print "\n\nINPUT4 = $INPUT4 \n \n ";

Answer 1

Have a read of perlrequick and perlretut.

The /e modifier of the s/// substitution operator treats the replacement as Perl code rather than as a string. For example:

$x = "5 10"
$x =~ s/(\d+) (\d+)/$1 + $2/e;
# $x is now 15

Instead of replacing $x with the string “$1 + $2”, it evaluates the Perl code $1 + $2 – where $1 is 5 and $2 is 10 – and puts the result into $x.

The || is not a regex operator, it’s a normal Perl operator. It is the logical-or operator: if the left-hand side is a true value (not 0 or ”), it returns the left side, otherwise it returns the right side. You can look up perl operators in perlop.