Concatinating variable names in bash script

Question

I am trying to run commands based on the output of a previous command:

ifaces=$(ls /sys/class/net)
for i in "${ifaces[@]}"
do
   mac=$(cat /sys/class/net/$i/address)
   echo "$i: $mac"
done

but its not working, i'm not sure if my method of concatenation is wrong?

Charles Duffy · Accepted Answer · 2018-04-17 20:45:12Z

for i in /sys/class/net/*; do
  mac=$(<"$i/address")
  echo "${i##*/}: $mac"
done

ls output is formatted for human readability rather than progammatic use. Because filenames can contain literal newlines, but ls separates names with newlines, it's impossible to both literally represent those names (without escape characters or sequences) and distinguish where on name ends and the next begins. Don't use it programatically.
$(<file), unlike $(cat "$file"), doesn't actually need to fork() off a subprocess, or exec() an external executable. Consequently, it's considerably more efficient.
${i##*/} removes the longest string matching the glob expression */ to the variable $i. Consequently, it effectively returns only the filename.

Evya · Accepted Answer · 2018-04-17 20:35:46Z

0

When your assigning the output of ls to ifaces, wrap it with an array to successfully iterate over the contents:

ifaces=($(ls /sys/class/net))
for i in "${ifaces[@]}"
do
   mac=$(cat /sys/class/net/$i/address)
   echo "$i: $mac"
done

answered Apr 17, 2018 at 20:35

Evya

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3 Comments

Yes, an array is the right thing. No, it shouldn't be populated from ls.

Duly noted, sir.

See BashPitfalls #1 as a reference.