dynamic variable assignment in shell script

Ok, let's grab bash manual and go:

export variable[=value] - "Mark each name to be passed to child processes in the environment.". That means that if you run another process from this shell (ie. a subshell with a command: sh -c 'echo $variable') it will share the variable value.

$* - "($*) Expands to the positional parameters, starting from one.". For example: sh -c 'echo $1' -- 1, $1 will expand to the "first position argument". So $* expands to all positional parameters.

${parameter:offset:length} - This is named "substring expansion". It expands to characters starting at offset. If the paramter is * or @ it exapnds to positional parameter, ie: ${*:2:1} is equal to $2. You can also give negative offset, it then counts from the back. If you omit :length part, it assumes that length is infinite (to the end of string, to the last positional parameter).

${*:-1} - this will not work as you expect, cause ${parameter:-word} expands to 1 if $* is null or unset. That means that we need a space between :- so bash does not confuse substring expansion with :-.

${*: -1} - this will give you the last positional parameter. ie. sh -c 'echo ${*: -1}' -- 1 2 3 will output 3.

export VAR_FILE=${*: -1} - get's the last positional parameter, assigns it to VAR_FILE variable and set's that VAR_FILE variable will be exported to subprocesses.

$* is a special variable which expands to the list of positional parameters (arguments to your script), separated by a space character.

${list: -1} expands to the last element of a list.

So this sets the environment variable VAR_FILE to the last argument passed to the script.

Note that for an array that you define yourself, the syntax would be different:

list=( a b 'c d' )
export foo=${list[*]: -1} # list[*] instead of just *