Bash - Perl Script variable insertion

Question

I'm having some troubles trying to insert a bash variable inside my perl code. The purpose of the code is to execute a certain perl printing command into a terminal using a bash script file. I have simplified my problem to the code below:

i=5
myvar=$(perl -e 'print "AAA" x ($i)')
echo "$myvar"

Having a variable i with 5 as its value, I want to print the string "AAA" as many times as the value of i, which is 5 in this example. To do this I want to use the given perl code. Currently, the perl code cannot read the content from my variable i. This is most likely due to the fact that " " is not being used for the part where I ask the value of i.

Switching the ' and the " is not really a posibility as the String "AAA" will be replaced by hexadecimal codes to be used for shellcode. Using the $ to execute the perl code is required to execute it in the terminal as an argument of a function. And of course, the value of i will increment/change in a loop so I cannot just use 5 instead of i.

Any help to make the perl code able to read the value of i or to work around this problem of the annotations?

Answer 1

Use this to enable variable expansion:

myvar=$(perl -e 'print "AAA" x ('"$i"')')

Answer 2

Perl has a rudimentary form of option parsing built-in. It works much like awk's -v option, but it's uglier

myvar=$(perl -s -e 'print "AAA" x ($j)' -- -j="$i")

That passes the shell's $i into perl as a perl variable $j. The double hyphen is required so that the -j is not gobbled up by the perl binary itself.

Answer 3

The problem is the type of quoting you're using. Single quotes cause shell variable expansion not to be done.

i=5
myvar=$(perl -e "print 'AAA' x ($i)")
echo "$myvar"

This code works as intended.