Modify global variable in while loop

In your example the while-loop is executed in a subshell, so changes to the variable inside the while-loop won't affect the external variable. This is because you're using the loop with a pipe, which automatically causes it to run in a subshell.

Here is an alternative solution using a while loop:

i=1
while read x; do
   i=$(($i + 1))
   echo $i
done <<<$(find tmp -type f)
echo $i

And here is the same approach using a for-loop:

i=1
for x in $(find tmp -type f);
do 
   i=$(($i + 1))
   echo $i
done
echo $i

For more information see the following posts:

  • A variable modified inside a while loop is not remembered

  • Bash Script: While-Loop Subshell Dilemma

Also look at the following chapter from the Advanced Bash Scripting Guide:

  • Chapter 23. Process Substitution

#!/bin/bash
i=1
while read x
do
   i=$((i+1))
   echo $i
done < <(find . -type f)
echo $i

https://stackoverflow.com/questions/7390497/bash-propagate-value-of-variable-to-outside-of-the-loop

Tags:

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