How to catch missing argument in function mistake? - linux

Question

I have just made the following mistake, where I am passing an argument to a function which is empty.
var1="ok"
var2=$notDefined
func $var1 $var2
func() {
var1=$1
var2=$2
echo $var1
echo $var2
}
For each argument in the function I could do
if [ -z $1 ]; then echo "Empty argument"; fi
But is there a more generic method to do this, so it is easy reusable, and would perhaps even tell the variable name that is empty?

Answer 1

You can stop whole script by set -u. It will fail if you try to use unset variable. It is very general approach.
Bash will output following localized message to standard error:
bash: x: unbound variable

Answer 2

You want to use the ? bash variable substitution operator:
var1=${1:?"undefined!"}
If $1 exists and isn't null, var1 is set to its value, otherwise bash prints 1 followed by "undefined!" and aborts the current command or script. This syntax can used for any bash variable.

Answer 3

In your case the empty variables are created, because there are too few arguments to the function.
You can get the number of passed arguments via $#. All variables that use $n with a higher number n must then be empty. You could check for a sufficiently high number of arguments at the beginning of your function.

Answer 4

#!/bin/bash
var1="ok"
var2=$notDefined
func() {
if [[ $# -ge 2 ]]; then
var1=$1
var2=$2
echo $var1
echo $var2
else
echo "Missing values"
fi
}
func $var1 $var2
Here it is running
./test.sh
Missing values
Here it is with two values:
#!/bin/bash
var1="ok"
var2="dokie"
func() {
if [[ $# -ge 2 ]]; then
var1=$1
var2=$2
echo $var1
echo $var2
else
echo "Missing values"
fi
}
func $var1 $var2
Results with:
./test.sh
ok
dokie