Linux shell scripting: Is it possible to modify the variable sent as parameter to a function?

Question

I am a bit new to linux shell scripting, so this could be a very silly question.

This is a simple example code (of course it does not work, but I am hoping to show what I wanna do):

#!/bin/bash 
function AppendLetters() {
  # Value sent as parameter: $1
  $1= "$1"LLL
}
var="foo"
AppendLetter $var
echo "$var"

So, when calling the program from command line:

$ ./example.sh

I would like to modify the internal variableobtain some sort of:

fooLLL

Reason for doing this: I have a script that loads multiple variables from a config file, and would like to make the same modification to that variables in order to use them in my program.

Is it possible? Can a function modify variables sent as parameter?

mywiki.wooledge.org/BashFAQ/006 covers all sorts of details about this sort of indirect assignment. — Etan Reisner
– Etan Reisner, Commented Dec 19, 2014 at 15:28

glenn jackman · Accepted Answer · 2014-12-19 16:48:34Z

2

Without needing bash 4.3, using variable indirection:

AppendLetters() {
  declare -g "$1"="${!1}LLL"
}

var=f00
AppendLetters var
echo "$var"

f00LLL

The -g option for declare is necessary so that the assignment is not treated as local to the function.

Given:

I have a script that loads multiple variables from a config file, and would like to make the same modification to that variables in order to use them in my program.

I would do this, not in a function, using bash's += assignment operator

varlist=( var1 var2 var3 )
for varname in "${varlist}"; do
    declare "$varname"+="LLL"
done

edited Dec 19, 2014 at 16:48

answered Dec 19, 2014 at 16:42

glenn jackman

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3 Comments

gniourf_gniourf Over a year ago

Same comment as in anubhava's post: you could definitely use declare -g "$1+=LLL" (this completely avoids indirection). Or use printf.

Sopalajo de Arrierez Over a year ago

Excuse me, @glennjackman, could you please tell me what does the ! symbol means in the ${!1}LLL part?

glenn jackman Over a year ago

Follow the "variable indirection" link at the top of my answer.

anubhava · Accepted Answer · 2014-12-19 18:35:55Z

2

Starting from Bash 4.3.0 you can use declare -n to get reference of a variable in a function:

AppendLetters() {
   declare -ln ref="$1"
   ref+="LLL"
}

var="foo"
AppendLetters "var"

echo "$var"
fooLLL

From help declare:

-n  make NAME a reference to the variable named by its value

EDIT: Thanks to @gniourf_gniourf for suggesting this printf, you can do this in older BASH:

AppendLetters() { printf -v "$1" '%sLLL' "${!1}"; }

var="foo"
AppendLetters "var"

echo "$var"
fooLLL

edited Dec 19, 2014 at 18:35

answered Dec 19, 2014 at 15:35

anubhava

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7 Comments

Sopalajo de Arrierez Over a year ago

Rulez! But is there any special reason for your { instead of my " in the appending LLL line?

anubhava Over a year ago

You could do ref="$ref"LLL also but this gives you flexibility to add a value with space also like: ref="${ref}LLL MM NN"

gniourf_gniourf Over a year ago

With Bash<4.3, you'd do: declare "$1"+=LLL. If you want to use declare -n (which is not really justified for this anyway, unless you want to explicitly check that you have a valid identifier—yet you could use printf for that), it's still better to use ref+=LLL. Also, declare your ref variable local as declare -ln ref="$1". Another option is to use printf with the -v switch.

gniourf_gniourf Over a year ago

You can use printf as: printf -v "$1" '%sLLL' "${!1}".

gniourf_gniourf Over a year ago

It's safer to use the command I gave: printf -v "$1" '%sLLL' "${!1}" in case the expansion of ${!1} yields percents or backslashes.

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