[bash] Bash integer comparison

I want to write a bash script that checks if there is at least one parameter and if there is one, if that parameter is either a 0 or a 1. this is the script:

#/bin/bash
if (("$#" < 1)) && ( (("$0" != 1)) ||  (("$0" -ne 0q)) ) ; then
echo this script requires a 1 or 0 as first parameter.
fi
xinput set-prop 12 "Device Enabled" $0

This gives the following errors:

./setTouchpadEnabled: line 2: ((: ./setTouchpadEnabled != 1: syntax error: operand expected (error token is "./setTouchpadEnabled != 1")
./setTouchpadEnabled: line 2: ((: ./setTouchpadEnabled -ne 0q: syntax error: operand expected (error token is "./setTouchpadEnabled -ne 0q")

What am I doing wrong?

The zeroth parameter of a shell command is the command itself (or sometimes the shell itself). You should be using $1.

(("$#" < 1)) && ( (("$1" != 1)) ||  (("$1" -ne 0q)) )

Your boolean logic is also a bit confused:

(( "$#" < 1 && # If the number of arguments is less than one…
  "$1" != 1 || "$1" -ne 0)) # …how can the first argument possibly be 1 or 0?

This is probably what you want:

(( "$#" )) && (( $1 == 1 || $1 == 0 )) # If true, there is at least one argument and its value is 0 or 1

Easier solution;

#/bin/bash
if (( ${1:-2} >= 2 )); then
    echo "First parameter must be 0 or 1"
fi
# rest of script...

Output

$ ./test 
First parameter must be 0 or 1
$ ./test 0
$ ./test 1
$ ./test 4
First parameter must be 0 or 1
$ ./test 2
First parameter must be 0 or 1

Explanation

• (( )) - Evaluates the expression using integers.
• ${1:-2} - Uses parameter expansion to set a value of 2 if undefined.
• >= 2 - True if the integer is greater than or equal to two 2.

I know this has been answered, but here's mine just because I think case is an under-appreciated tool. (Maybe because people think it is slow, but it's at least as fast as an if, sometimes faster.)

case "$1" in
    0|1) xinput set-prop 12 "Device Enabled" $1 ;;
      *) echo "This script requires a 1 or 0 as first parameter." ;;
esac