Question

Is there a way in bash to understand the name of a variable that’s passed to it?

Example:

var1=file1.txt
err_var=errorfile.txt

function func1 {
   echo "func1: name of the variable is: " $1
   echo "func1: value of variable is: " $1 
   echo
   if [ ! -e $var1 ]
   then
      $1 = $err_val  #is this even possible?
   fi
   func2 $1
}

function func2 {
   echo "func2: name of the variable is: " $1
   echo "func2: value of variable is: " $1 
   echo
}

func1 $var1
func1 $err_var

I was hoping to get the following output if file1.txt exists:

func1: name of the variable is: var1
func1: value of variable is: file1.txt

func2: name of the variable is: var1
func2: value of variable is: file1.txt

And when file1.txt does not exist:

func1: name of the variable is: var1
func1: value of variable is: file1.txt

func2: name of the variable is: err_var
func2: value of variable is: errorfile.txt

Any ideas?

Answer 1

No, the variable is expanded before the function sees it. The function only sees the value, not the variable name.

If you pass the variable name unexpanded and without the dollar sign, you can use indirection.

get_it () {
    echo "${!1}"
}

Demo:

$ foo=bar
$ baz=qux
$ get_it foo
bar
$ get_it baz
qux