Variable scope for bash shell scripts and functions in the script

Question

I'm a little confused with my script regarding functions, variable scope, and possibly subshells. I saw in another post that pipes spawn a subshell and the parent shell can't access variables from the subshell. Is this the same case with cmds run in backticks too?

To not bore people, I've shortened my 100+ line script but I tried to remember to leave in the important elements (i.e. backticks, pipes etc). Hopefully I didn't leave anything out.

global1=0
global2=0
start_read=true

function testfunc {
   global1=9999
   global2=1111
   echo "in testfunc"
   echo $global1
   echo $global2
}

file1=whocares
file2=whocares2

for line in `cat $file1`
do
   for i in `grep -P "\w+ stream" $file2 | grep "$line"`   # possible but unlikely problem spot
   do
         end=$(echo $i | cut -d ' ' -f 1-4 | cut -d ',' -f 1)   # possible but unlikely spot
         duration=`testfunc $end`       # more likely problem spot
   done
done

echo "global1 = $global1"
echo "global2 = $global2"

So when I run my script, the last line says global1 = 0. However, in my function testfunc, global1 gets set to 9999 and the debug msgs print out that within the function at least, it is 9999.

Two questions here:

1. Do the backticks spawn a subshell and thus making my script not work?
2. How do I work around this issue?

Answer 1

You can try something like

global1=0
global2=0
start_read=true

function testfunc {
   global1=9999
   global2=1111
   echo "in testfunc"
   echo $global1
   echo $global2
   duration=something
}

file1=whocares
file2=whocares2

for line in `cat $file1`
do
   for i in `grep -P "\w+ stream" $file2 | grep "$line"`   # possible but unlikely problem spot
   do
         end=$(echo $i | cut -d ' ' -f 1-4 | cut -d ',' -f 1)   # possible but unlikely spot
         testfunc $end       # more likely problem spot
   done
done

echo "global1 = $global1"
echo "global2 = $global2"