Bash integer comparison

Solution 1:

This script works!

#/bin/bash
if [[ ( "$#" < 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" $0
fi

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/bash
if (( $# )) && (( $1 == 0 || $1 == 1 )); then
    echo "first parameter is $1"
    xinput set-prop 12 "Device Enabled" $0
else
    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 0
first parameter is 0

$ ./tmp.sh 1
first parameter is 1

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

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

Solution 2:

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.

Solution 3:

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