how can I use a variable as an array name in bash

Question

I am currently working on a script that receives input from stdin, and then loads the input line by line into an array for further processing. Although the array works fine if I set the array name directly, I cannot get the array part to work properly when I attempt to use a variable as the array name, the code itself is below:

input=$(</dev/stdin)
# back up the field separator for later
OLDIFS=$IFS
# set the field separator to newline
IFS=$'\n'
# populate an array from that variable, as delimited by the IFS
lines=($input)

and this is what I have tried for setting the array name as a variable

arrayname="something"
eval $arrayname=($input)

but unfortunately when I go to run this, I get the following error:

./f.sh: line 53: syntax error near unexpected token `('
./f.sh: line 53: `                      eval $arrayname=($input)'

having said all of this, I was wondering if anyone would know what I could do to make this work correctly? thanks!

gniourf_gniourf · Accepted Answer · 2012-11-30 22:56:30Z

3

Instead of playing with IFS, which is useless in your case (even though I'm an avid fan of IFS, I tend to only use it when necessary!), there's a much better and much more efficient way of loading a file into an array. Incidentally, it is not very well-known: it's the command mapfile (type help mapfile in your terminal running bash for more info).

Hence,

mapfile lines

will read from standard input and store each line in an array lines. This also stores the trailing new line character, so if you want to trim it, use the -t option:

mapfile -t lines

To print out the array, one line per entry, use:

printf '%s\n' "${lines[@]}"

(observe the quoting for "${lines[@]}").

Please note that using eval is evil!

Your main problem is to use a variable as an array name. bash doesn't have what is called a pointer in other languages. Without using eval, it will be difficult to use an array name stored in a variable (unless you use little trickeries). But, as I already said, eval is evil. Maybe you should spend a little more time thinking about your design, there's probably another (better?) way to achieve what you're trying to.

edited Nov 30, 2012 at 22:56

answered Nov 30, 2012 at 21:54

gniourf_gniourf

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2 Comments

lacrosse1991 Over a year ago

thanks for the info! after what you had mentioned with thinking about the design, I've thought of a way of not having to use a variable as a name for the array, although it would require that I append data onto an array instead, do you know if that would be possible? for instance I move one file into an array (say it contained 40 lines or so), after which I would add a second file onto the array

lacrosse1991 Over a year ago

was able to figure it out, thanks again for your help though :)

choroba · Accepted Answer · 2012-11-30 22:48:28Z

1

Use indirection:

men=(Jack John Joe)
women=(Zoe Zelda Zanna)
gender=women[@]
echo ${!gender}

answered Nov 30, 2012 at 22:48

choroba

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Comments

Gilles Quénot · Accepted Answer · 2012-11-30 21:44:52Z

0

read -a arr
printf '%s\n' ${arr[@]}

Is it the solution ?

answered Nov 30, 2012 at 21:44

Gilles Quénot

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