Question

I have this Perl script:

#!/usr/bin/perl

$var = `ls -l \$ddd` ;
print $var, "\n";

And ddd is a shell variable

$ echo "$ddd"
arraytest.pl

When I execute the Perl script I get a listing of all files in the directory instead of just one file, whose file name is contained in shell variable $ddd.

Whats happening here ? Note that I am escaping $ddd in backticks in the Perl script.

Answer 1

The variable $ddd isn’t set *in the shell that you invoke from your Perl script.

Ordinary shell variables are not inherited by subprocesses. Environment variables are.

If you want this to work, you’ll need to do one of the following in your shell before invoking your Perl script:

ddd=arraytest.pl ; export ddd # sh

export ddd=arraytest.pl       # bash, ksh, zsh

setenv ddd arraytest.pl       # csh, tcsh

This will make the environment variable $ddd visible from your Perl script. But then it probably makes more sense to refer to it as $ENV{ddd}, rather than passing the literal string '$ddd' to the shell and letting it expand it:

$var = `ls -l $ENV{ddd}`;