fix scope problem with variable in bash script

26 Jul 2022

      Obviously I've reduced and rewitten it to better point out the problem 
in this script.  The question is, how should the variable "count" be 
declared so that it isn't treated as two, completely different variables 
(with exactly the same name)?

If you want to try to run it, it takes a filename as an commandline 
argument.

========================
#!/bin/bash

#How to declare the variable "count"???
count=0

cat $1 | while read line
do
     count=$((count+1))
     echo -n "$count "
done

echo
echo File $1 has $count lines.

========================

Thanks.

kf

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