Answer by fozzybear for Can command substitution be nested in variable substitution?

Here's an examplary hack, that bypasses assigning a temporary variable for chained parameter substitution:

declare -a values=( "4: " "#1" "#2" "#3" "#4" )declare -A arr["VALUES"]="${values[@]}"echo -e "Original: ${arr[@]}""${arr[@]#*: ?}" >&/dev/null; san="${_//#/}"echo -e "Sanitized: $san"

Explanation:

We initialize an associative array arr with values from array values.The first element contains the number of value elements, followed by a :; this needs to be filtered out, as well as the prefix # for all values.

"${arr[@]#*: ?}" >&/dev/null

This is a parameter substitution, that first expands the values of arr and filters everything in front, up to and including '4: ' (#*: ?), discarding the result, both of stdout and stderr. This also suppresses an error, too, due to the missing variable assigning of the result.

>& is a short form of 2>&1, assigning the stderr output (fd2) to stdout (fd1) and both to /dev/null.

The trick now is, to use bash's special underscore _ parameter, which copies the discarded result parameter from the last command (1st parameter expansion operation above and result redirection to /dev/null) into our second parameter substitution, where the # value prefixes are filtered:

san="${_//#/}"

Output:

Original: 4: #1 #2 #3 #4

Sanitized: 1 2 3 4

Also see: Special parameters and shell variables