Shell Programming and Scripting

I would like to remove comments from a bash script. In addition, I would like to remove lines that consist of only white spaces, and to remove blank lines.

Code:

#!/bin/bash
perl -pe 's/ *#.*$//g' $1 | grep -v ^[[:space:]]*$ | perl -pe 's/  +/ /g' > $2
#
# $1 INFILE
# $2 OUTFILE

The above code seemed to work at first. Unfortunately, however, I found later that the above code destroys the following two special variables.

${#ARRAY[@]} the number of array elements
$# the number of shell arguments

A workaround is to replace "${#" and "$#" with words that do not appear in the input file before applying the above code.

Code:

sed 's/${#/__UNUSUALWORD1__/g ; s/$#/__UNUSUALWORD2__/g' in.txt | \
perl -pe 's/ *#.*$//g' | grep -v ^[[:space:]]*$ | perl -pe 's/  +/ /g' | \
sed 's/__UNUSUALWORD1__/${#/ ; s/__UNUSUALWORD2__/$#/' > out.txt

However, the preparatory replacement is awkward. I would like to modify 's/ *#.*$//g' so that it will not match "${#" or "$#". Does anyone know a better solution?

Bash comments always start with #. However, the problem is that bash allows some exceptions where # does not lead a comment, as shown below.

Code:

${#ARRAY[@]} the number of array elements
$# the number of shell arguments
\# escaped by a backslash.
'abcd#efgh' protected by quotes
"abcd#efgh" protected by quotes

Does anyone know how to remove comments from bash scripts without destroying the exempted #'s that do not lead comments? (In addition, I would like to remove lines that consist of only white spaces, and to remove blank lines.)

Many thanks in advance.