cat command outputs the whole directory when used in a bash script

Question

I have a scirpt like this:

#!/bash/bin
file=`cat /proc/net/arp`
echo $file

And when I run the script it outputs the whole folder where the script is like this:

IP address HW type Flags HW address Mask Device 192.168.1.102 0x1 0x2 a0:88:b4:e3:f6:6c naloga1.png naloga1.txt naloga1.txt~ naloga2.png naloga2.txt naloga2.txt~ naloga3.png naloga3.sh naloga3.sh~ naloga3.txt naloga3.txt~ naloga4.png naloga4.txt naloga5.png naloga5.sh naloga5.sh~ naloga5.txt naloga6.png naloga6.txt naloga6.txt~ naloga7.png naloga7.txt naloga7.txt~ naloga8.sh naloga8.sh~ porocilo.odt tekst.txt tekst.txt~ eth0 192.168.1.1 0x1 0x2 58:6d:8f:15:fd:c3 naloga1.png naloga1.txt naloga1.txt~ naloga2.png naloga2.txt naloga2.txt~ naloga3.png naloga3.sh naloga3.sh~ naloga3.txt naloga3.txt~ naloga4.png naloga4.txt naloga5.png naloga5.sh naloga5.sh~ naloga5.txt naloga6.png naloga6.txt naloga6.txt~ naloga7.png naloga7.txt naloga7.txt~ naloga8.sh naloga8.sh~ porocilo.odt tekst.txt tekst.txt~ eth0

Any idea why it is acting this way?

Answer 1

The file contains a wildcard (probably *), and when you echo it you don't have double-quotes around the $file variable reference. When the shell expands a variable reference like that, it goes back through the value and parses a number of things: spaces (treated as separators between arguments), wildcards (which are replaced by the appropriate list of filenames), etc. Putting the variable reference inside double-quotes avoids this extra parsing. In general, it's a good idea to put all variable references in double-quotes unless you specifically want the extra parsing for some reason.

BTW, I also prefer $() over backquotes -- they do essentially the same thing, but the parenthesis version is much easier to read. Also, in this context bash will let you skip the cat command and just use a < redirection (note that this is a bash-only feature, though).

#!/bash/bin
file=$(</proc/net/arp)
echo "$file"