Shell variables set inside while loop not visible outside of it

Question

I am trying to find the pathname with the most characters in it. There might be better ways to do this. But I would like to know why this problem occurs.

LONGEST_CNT=0
find samples/ | while read line
do
    line_length=$(echo $line | wc -m)
    if [[ $line_length -gt $LONGEST_CNT ]] 
    then
        LONGEST_CNT=$line_length
        LONGEST_STR=$line
    fi
done

echo $LONGEST_CNT : $LONGEST_STR

It somehow always returns:

0 :

If I print the results for debugging inside the while loop the values are correct. So why bash does not make these variables global?

This is (and was when asked) BashFAQ #24

Charles Duffy
– Charles Duffy

2025-03-11 16:12:29 +00:00
Commented Mar 11 at 16:12 — Charles Duffy
– Charles Duffy, Commented Mar 11 at 16:12

Dennis Williamson · Accepted Answer · 2012-02-29 16:11:24Z

89

When you pipe into a while loop in Bash, it creates a subshell. When the subshell exits, all variables return to their previous values (which may be null or unset). This can be prevented by using process substitution.

LONGEST_CNT=0
while read -r line
do
    line_length=${#line}
    if (( line_length > LONGEST_CNT ))
    then
        LONGEST_CNT=$line_length
        LONGEST_STR=$line
    fi
done < <(find samples/ )    # process substitution

echo $LONGEST_CNT : $LONGEST_STR

edited Feb 29, 2012 at 16:11

answered Jan 12, 2011 at 10:30

Dennis Williamson

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5 Comments

RobSis Over a year ago

The solution Dennis provided works, but be aware, that it violates POSIX norm. Try to 'set -o posix' and the script will not work!

Dennis Williamson Over a year ago

@Robert: The question is tagged [bash]. It is true that process substitution is not specified by POSIX. However, unless portability to POSIX-only shells is a concern, there's no reason not to use Bash-specific features. By the way, process substitution is supported by ksh and zsh. However, they don't create a subshell that causes the variable to be lost. Also note that Bash 4.2 has an option, shopt -s lastpipe which "runs the last command of a pipeline in the current shell context" (not a subshell) - unless job control is in affect.

Aswin Murugesh Over a year ago

I use this, but get syntax error: syntax error near unexpected token done'`

Dennis Williamson Over a year ago

@AswinMurugesh: Most likely, you are missing the do.

Alois Mahdal Over a year ago

return to their previous values ... well, IMHO better formulation would be that the changes happen in a temporary copy of environment (even different PID I think) that gets destroyed as the while loop is over. The original global variable is never touched at all; there's no reason to "return".

mivk · Accepted Answer · 2025-03-11 15:57:07Z

21

The "correct" reply is given by Dennis. However, I find the process substitution trick extremely unreadable if the loop contains more than a few lines. When reading a script, I want to see what goes into the pipe before I see how it is processed.

So I usually prefer this trick of encapsulating the while loop in "{}".

LONGEST_CNT=0
find /usr/share/zoneinfo | \
{ while read -r line
    do
        line_length=${#line}
        if (( line_length > LONGEST_CNT ))
        then
            LONGEST_CNT=$line_length
            LONGEST_STR=$line
        fi
    done
    echo $LONGEST_CNT : $LONGEST_STR
}

This makes the variable available after the loop, but of course only while still inside the command grouping, so before the closing brace.

edited Mar 11 at 15:57

answered Dec 2, 2012 at 1:16

mivk

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4 Comments

0xC0000022L Over a year ago

Totally agree about the readability point. +1. Still, it's the way to go.

Deim0s Over a year ago

except that this example does not modify the global variable. if you check LONGEST_CNT after the loop, you will see it's still zero

mivk Over a year ago

@Deims0s : yes. That is why the echo statement is inside the "{}". If the variables are needed again later, the process substitution way might be better suited.

Sildoreth Over a year ago

This doesn't solve the problem. Readability counts for nothing if the code doesn't achieve its goal.

Alois Mahdal · Accepted Answer · 2015-12-03 19:10:38Z

2

Do what you always (should) do:

separate concerns,
avoid globals,
document your code,
be readable,
maybe be POSIXy.

(Yes, I added a bit more "good practice" ingredients to the soup than absolutely necessary ;))

So my favorite "knee-jerk reaction" to problems with invisible subshell is to use function:

#!/bin/sh

longest() {
    #
    # Print length and body of the longest line in STDIN
    #
    local cur_ln    # current line
    local cur_sz    # current size (line length)
    local max_sz    # greatest size so far
    local winner    # longest string so far
    max_sz=0
    while read -r cur_ln
    do
        cur_sz=${#cur_ln}
        if test "$cur_sz" -gt "$max_sz";
        then
            max_sz=$cur_sz
            winner=$cur_ln
        fi
    done
    echo "$max_sz" : "$winner"
}

find /usr/share/zoneinfo | longest

# ok, if you really wish to use globals, here you go ;)
LONGEST_CNT=0
LONGEST_CNT=$(
    find /usr/share/zoneinfo \
      | longest \
      | cut -d: -f1 \
      | xargs echo\
)
echo "LONGEST_CNT='$LONGEST_CNT'"

Aside from avoiding the subshell annoyance, it gives you perfect place to document the code, and sort-of adds namespacing: notice that inside function you can use much shorter and simpler variable names without losing any of readability.

answered Dec 3, 2015 at 19:10

Alois Mahdal

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2 Comments

Priv Acyplease Over a year ago

The last part of this is the best answer really, but fyi you can put the pipes at the ends of the lines and avoid the line continuation backslashes.

Alois Mahdal Over a year ago

@PrivAcyplease the intentional advantage of this style is that it makes the pipeline structure obvious, especially for commands in the middle. That's IMO well worth the extra chars.

ruakh · Accepted Answer · 2025-03-11 16:13:33Z

2

About finding the longest pathname. Here's an alternative:

find /usr/share/zoneinfo | while read line; do
    echo "${#line} $line" 
done | sort -nr | head -n 1

# Result:
58 /usr/share/zoneinfo/right/America/Argentina/ComodRivadavia

Forgive me if this is considered off topic, I hope it helps someone.

edited Mar 11 at 16:13

ruakh

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answered Dec 27, 2013 at 9:08

grebneke

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6 Comments

tripleee Over a year ago

Using echo to print the output from backicks is rarely useful or necessary. See useless use of echo.

grebneke Over a year ago

@tripleee: AFAICT both echos are useful in this example: wc needs the input from $line, and $line needs to be displayed after the output from wc. Please, feel free to suggest improvements without adding more lines or variables.

Anthony Geoghegan Over a year ago

@grebneke I upvoted your answer (and comment) but after thinking about it some more, I realised you could avoid using command substitution to run wc. Instead, you could use POSIX shell parameter expansion: echo ${#line} $line;

grebneke Over a year ago

@AnthonyGeoghegan nice suggestion. To be clear, your example uses # for String Length as opposed to Number of positional parameters - I think you accidentally posted the wrong link. # is used for both and it can get confusing.

grebneke Over a year ago

@AnthonyGeoghegan updated with your solution, way faster, gj tyvm

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