Executing the output as filename

Question

In one of my Bash scripts, there's a point where I have a variable SCRIPT which contains the /path/to/an/exe, and what the script ultimately needs to do, is executing that executable. Therefore the last line of the script is

$($SCRIPT)

so that $SCRIPT is expanded to /path/to/an/exe, and $(/path/to/an/exe) executes the executable.

However, running shellcheck on the script generates this error:

In setscreens.sh line 7:
$($SCRIPT)
^--------^ SC2091: Remove surrounding $() to avoid executing output.

For more information:
  https://www.shellcheck.net/wiki/SC2091 -- Remove surrounding $() to avoid e...

Is there a way I can rewrite that $($SCRIPT) in a more appropriate way? eval does not seem to be of much help here.

Answer 1

$($SCRIPT) indeed does not do what you think it does.

The outer $() will execute any commands inside the parenthesis and execute the result string.

The inner $SCRIPT will expand to the value of the SCRIPT variable and execute this string while splitting words on spaces/

If you want to execute the command contained into the SCRIPT variable, you just write as an example:

SCRIPT='/bin/ls'
"$SCRIPT" # Will execute /bin/ls

Now if you also need to handle arguments with your SCRIPT variable command call:

SCRIPT='/bin/ls'
"$SCRIPT" -l # Will execute /bin/ls -l

To also store or build arguments dynamically, you'd need an array instead of a string variable.

Example:

SCRIPT=(/bin/ls -l)
"${SCRIPT[@]}" # Will execute /bin/ls -l

SCRIPT+=(/etc) # Add /etc to the array
"${SCRIPT[@]}" # Will execute /bin/ls -l /etc