Scope of global variable in shell script [duplicate]

Question

I have a shell script like this , i am reading a file and counting the line number. i got logs like 1,2,3,4.. etc from inside processsNumber function

#!/bin/bash

number=0;
processNumber () {
    ((number++));
    echo "$number";
}

grep -E '*' readme.txt | while read -r line ; do
    processNumber "$line";
done;
echo "And at last $number";

But it logs "And at last 0" , but i was expecting the last line number. Why its like that ? is it because gres reads a file a assync call hence echo has to be a callback to that.

Or is that global variable changes cannot be tracked outside the function.?

How to fix this with out changing the grep and pipe ?

NOTE: my target is not find the number of lines in the file , but to understand this

anubhava · Accepted Answer · 2014-09-16 10:57:02Z

4

But it logs "And at last 0"

That is because you're calling processNumber function using a pipe which makes it execute in a sub shell not in the main shell hence in parent shell variable remains unchanged to 0.

UPDATE: To avoid creating pipe (and subshell) use for loop like this:

while read -r line; do
    processNumber "$line"
done < readme.txt

edited Sep 16, 2014 at 10:57

answered Sep 16, 2014 at 10:47

anubhava

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7 Comments

Sarath Over a year ago

is there a way to make callbacks in shell like javascriot callbacks ?

anubhava Over a year ago

Shell doesn't have callback like JS has in String#replace but if you clarify what you want to do we can sure help. But I also request to not to change the question completely in doing do.

Sarath Over a year ago

i need a method to how to fix this, without changing the pipe logic

anubhava Over a year ago

ok check my updated answer for that.

anubhava Over a year ago

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