20

I want to take the absolute of a number by the following code in bash:

#!/bin/bash
echo "Enter the first file name: "
read first

echo "Enter the second file name: "
read second

s1=$(stat --format=%s "$first")
s2=$(stat -c '%s' "$second")

res= expr $s2 - $s1

if [ "$res" -lt 0 ]
then
        res=$res \* -1
fi

echo $res

Now the problem I am facing is in the if statement, no matter what I changes it always goes in the if, I tried to put [[ ]] around the statement but nothing.

Here is the error:

./p6.sh: line 13: [: : integer expression expected
John Kugelman
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Ali Sajid
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  • one of the easier ways could be adding extra line after calculating the $res res=`echo $res | tr -d -` – prash Sep 25 '17 at 11:25

4 Answers4

44

You might just take ${var#-}.

${var#Pattern} Remove from $var the shortest part of $Pattern that matches the front end of $var. tdlp

Example:

s2=5; s1=4
s3=$((s1-s2))

echo $s3
-1

echo ${s3#-}
1
Suuuehgi
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7
$ s2=5 s1=4
$ echo $s2 $s1
5 4
$ res= expr $s2 - $s1
1
$ echo $res

What's actually happening on the fourth line is that res is being set to nothing and exported for the expr command. Thus, when you run [ "$res" -lt 0 ] res is expanding to nothing and you see the error.

You could just use an arithmetic expression:

$ (( res=s2-s1 ))
$ echo $res
1

Arithmetic context guarantees the result will be an integer, so even if all your terms are undefined to begin with, you will get an integer result (namely zero).

$ (( res = whoknows - whocares )); echo $res
0

Alternatively, you can tell the shell that res is an integer by declaring it as such:

$ declare -i res
$ res=s2-s1

The interesting thing here is that the right hand side of an assignment is treated in arithmetic context, so you don't need the $ for the expansions.

kojiro
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    @AliSajid no, the _error_ is at `if`. The _problem_ is before `if`, when you never actually assign a value to `res`. – kojiro Mar 24 '15 at 01:47
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    abs $res => `${res/#-/}` :) – rici Mar 24 '15 at 02:02
  • @rici heh. I thought of that, but ruled it out because OP said _by the following code_. But yeah… – kojiro Mar 24 '15 at 02:08
  • @kojiro `$ s2=5 s1=4` `$ echo $s2 $s1` `5 4` `$ res= expr $s1 - $s2` `-1` `$ echo $res` `{empty line}` – Vitalii Diravka Oct 08 '19 at 08:53
  • Sorry dude) Looks like you answer is outdated. Anyway rici answer helped me: `${VAR1#-}` works fine – Vitalii Diravka Oct 09 '19 at 17:40
  • Nah, my answer is not outdated. It's perfectly fine and helps OP understand the error they were getting. The _logic_ of OP's solution is perfectly fine for getting an absolute value. There are other solutions and that's cool too. – kojiro Oct 09 '19 at 20:24
7

I know this thread is WAY old at this point, but I wanted to share a function I wrote that could help with this:

abs() { 
    [[ $[ $@ ] -lt 0 ]] && echo "$[ ($@) * -1 ]" || echo "$[ $@ ]"
}

This will take any mathematical/numeric expression as an argument and return the absolute value. For instance: abs -4 => 4 or abs 5-8 => 3

bng44270
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1

A workaround: try to eliminate the minus sign.

  1. with sed
x=-12
x=$( sed "s/-//" <<< $x )
echo $x

12
  1. Checking the first character with parameter expansion
x=-12
[[ ${x:0:1} = '-' ]] && x=${x:1} || :
echo $x

12

This syntax is a ternary opeartor. The colon ':' is the do-nothing instruction.

  1. or substitute the '-' sign with nothing (again parameter expansion)
x=-12
echo ${x/-/}

12

Personally, scripting bash appears easier to me when I think string-first.

MarcoP
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