Syntax error in subtraction in Bash


 
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# 1  
Old 10-21-2019
Syntax error in subtraction in Bash

I am sharing a code snippet.
Code:
for (( i=0; i<=$(( $count -1 )); i++ ))
do

	first=${barr2[$i]}
	search=${barr1[$i]}
        echo $first
	echo "loop begins"
	for (( j=0; j<=5000; j++ ))
	do
		if [[ ${harr1[j]} == $search ]]; then
			echo $j
			break;
		fi

	done
	second=${harr2[$j]}	
	echo $second
	
	diff=$((second-first))
	echo "diff" 
	echo $diff
done

I am getting a syntax error in line of subtraction that says something like:
syntax error: invalid arithmetic operator (error token is ".0985146")

This is the line that gives error.
Code:
	diff=$((second-first))

Appreciate your help.
# 2  
Old 10-21-2019
I am assuming you are using bash or perhaps a POSIX shell like dash.
Your error is telling you that it is not possible to do such a task as these shells have INTEGER arithmetic only.

If you need floating/fixed point arithmetic then you will have to do workarounds using, bc, dc, python, perl, awk or any other methods that are capable of such mechanisms.

(Just an observation, you have both, ${barr1[$i]} and also ${harr1[j]} are these correct?)

EDIT:
Note the second one should have ${harr1[$j]}

An example longhand using INTEGER arithmetic and FIXED point, OSX 10.14.6, default bash terminal:
Code:
Last login: Mon Oct 21 18:25:40 on ttys000
AMIGA:amiga~> x=.098765
AMIGA:amiga~> y=10.3
AMIGA:amiga~> x=$( printf "%.f" ${x}e+12 )
AMIGA:amiga~> y=$( printf "%.f" ${y}e+12 )
AMIGA:amiga~> printf "%.12f\n" $(( y - x ))e-12
10.201235000000
AMIGA:amiga~> _


Last edited by wisecracker; 10-21-2019 at 02:50 PM.. Reason: See EDIT:
These 3 Users Gave Thanks to wisecracker For This Post:
# 3  
Old 10-21-2019
Nice wisecracker. This method would rap up into a fun little bash sum function:

Code:
function sum
{
   local word exp var

   [[ "$1" = *= ]] && var="-v ${1%=}" && shift

   for word in "$@"
   do
       case "$word" in
          [0-9.]*|-[0-9.]*|+|-)
              printf -v ans "%.f" "${word}e+12" 2>/dev/null || ans=$word
              exp="$exp $ans" ;;
          *) echo "sum: $word operator not supported" >&2 ; exit 1 ;;
       esac
   done
   printf $var "%.8f\n" $(( $exp ))e-12
}

sum val= 3.58047 - .68 + .858 + -1.2
echo $val

These 2 Users Gave Thanks to Chubler_XL For This Post:
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