mjhjmtu

I found a script that has a function that checks if a variable is set but i don't understand it very well.

check_if_variable_is_set() 
 var_name=$1
 if [ -z "$!var_name+x" ]; then
 false
 else
 true
 fi

What exactly happens with this substitution ?

In the bash shell, $!var is a variable indirection. It expands to the value of the variable whose name is kept in $var.

The variable expansion $var+value is a POSIX expansion that expands to value if the variable var is set (no matter if its value is empty or not).

Combining these, $!var+x would expand to x if the variable whose name is kept in $var is set.

Example:

$ foo=hello
$ var=foo

$ echo "$!var+$var is set, its value is $!var"
foo is set, its value is hello

$ unset foo
$ echo "$!var+$var is set, its value is $!var"

(empty line as output)

The function in the question could be shortened to

check_if_variable_is_set () [ -n "$!1+x" ]; 

or even:

check_if_variable_is_set () [ -v "$1" ]; 

or even:

check_if_variable_is_set()[[ -v $1 ]]

Where -v is a bash test on a variable name which will be true if the named variable is set, and false otherwise.

POSIXly, it could be written:

check_if_variable_is_set() eval '[ -n "$'"$1"'+x" ]'; 

Note that all those are potential command injection vulnerabilities if the argument to that function could end up being under control of an attacker. Try for instance with check_if_variable_is_set 'a[$(id>&2)]'.

To guard against that, you may want to verify that the argument is a valid variable name first. For variables:

check_if_variable_is_set() [![:alpha:]_]*) false;;
 (*) eval '[ -n "$'"$1"'+x" ]'
 esac

(note that [[:alpha:]] will check for alphabetical characters in your locale while some shells only accept alphabetical characters from the portable character set in their variable)