Bash integer comparison

Question

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?

It looks like you are running your script using sh ./setTouchpadEnabled instead of using bash. — jordanm
– jordanm, Commented Jan 25, 2013 at 5:35
@jordanm are you referring to the lack of a bang in the shebang line? — Kev
– Kev, Commented Oct 18, 2017 at 13:06

Matthias Braun · Accepted Answer · 2017-11-13 17:16:17Z

56

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}

edited Nov 13, 2017 at 17:16

Matthias Braun

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answered Jan 24, 2013 at 22:05

user000001

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4 Comments

Cheiron Over a year ago

Thank you very much, the error is because the command is xinput and not input

0xC0000022L Over a year ago

Any particular reason you didn't stick with the arithmetic expressions also used in the question? I.e. use of (( and )).

0xC0000022L Over a year ago

@user828193: well clearly this is about calculations, so arithmetic expressions are the way to go. Which is why I found it irritating that you changed this aspect of the question in your answer.

user000001 Over a year ago

@0xC0000022L: You are right! Here it is with arithmetic expressions. I guess I got carried off by the -ne and the comparison to a possibly empty string in the OP's code...

koola · Accepted Answer · 2013-01-25 05:10:23Z

20

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.

answered Jan 25, 2013 at 5:10

koola

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1 Comment

SpinUp __ A Davis Over a year ago

+1, this exposes and works around a not-great behaviour of the (( command, which assigns the value of 0 to an empty or unset variable.

kojiro · Accepted Answer · 2013-01-24 21:59:46Z

9

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

edited Jan 24, 2013 at 21:59

answered Jan 24, 2013 at 21:42

kojiro

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4 Comments

Cheiron Over a year ago

This is an improvement, thank you. It however still gives me errors: ./setTouchpadEnabled: line 2: ((: != 1: syntax error: operand expected (error token is "!= 1") ./setTouchpadEnabled: line 2: ((: != 0: syntax error: operand expected (error token is "!= 0")

kojiro Over a year ago

The quotes aren't strictly necessary inside (( )), but they make StackOverflow's highlighting smarter.

kojiro Over a year ago

@Cheiron I think you misunderstood my comment. Take another look at my answer.

Cheiron Over a year ago

Oh wow, I was indeed doing boolean logic wrong. It still gives me errors, though: ./setTouchpadEnabled: line 2: ((: 0 && ( == 1 || == 0 ) : syntax error: operand expected (error token is "== 1 || == 0 ) ")

William · Accepted Answer · 2013-01-25 15:03:30Z

8

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

answered Jan 25, 2013 at 15:03

William

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1 Comment

SpinUp __ A Davis Over a year ago

That is nice and clean. Note also the evaluated variable in a case statement is one of those places where you can safely leave it unquoted.

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