Bash integer comparison

This script works!

#/bin/bashif [[ ( "$#" < 1 ) || ( !( "$1" == 1 ) && !( "$1" == 0 ) ) ]] ; then    echo this script requires a 1 or 0 as first parameter.else    echo "first parameter is $1"    xinput set-prop 12 "Device Enabled" $0fi

But this also works, and in addition keeps the logic of the OP, since the question is about calculations. Here it is with only arithmetic expressions:

#/bin/bashif (( $# )) && (( $1 == 0 || $1 == 1 )); then    echo "first parameter is $1"    xinput set-prop 12 "Device Enabled" $0else    echo this script requires a 1 or 0 as first parameter.fi

The output is the same¹:

$ ./tmp.sh this script requires a 1 or 0 as first parameter.$ ./tmp.sh 0first parameter is 0$ ./tmp.sh 1first parameter is 1$ ./tmp.sh 2this script requires a 1 or 0 as first parameter.

_{[1] the second fails if the first argument is a string}

Easier solution;

#/bin/bashif (( ${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 4First parameter must be 0 or 1$ ./test 2First 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.

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