How to perform arithmetic in UNIX?

Question

I am trying to build a very basic 4 function arithmetic script in UNIX and it doesn't like my arithmetic statement. I was trying to use 'bash arithmetic' syntax

from this source

http://faculty.salina.k-state.edu/tim/unix_sg/bash/math.html

also as a side note when you reference a variable in UNIX when do you need to use the "$" symbol, and when do you not need to?

#!/bin/bash

str1="add"
str2="sub"
str3="div"
str4="mult"

((int3=0))
((int2=0))
((int1=0))


clear
read -p "please enter the first integer" $int1
clear
read -p "Please enter mathmatical operator'" input
clear
read -p "Please enter the second integer" $int2


if [ "$input" = "$str1" ];
then

((int3 = int1+int2))
   echo "$int3"


else

   echo "sadly, it does not work"

fi;

Possible duplicate of [How can I add numbers in a bash script](http://stackoverflow.com/questions/6348902/how-can-i-add-numbers-in-a-bash-script) — l'L'l, Feb 20 '17 at 18:11

score 1 · Answer 1 · answered Jun 29 '17 at 02:36

I think this is what you want:

#!/bin/bash

str1="add"
str2="sub"
str3="div"
str4="mult"

((int3=0)) # maybe you can explain me in comments why you need a arithmetic expression here to perform an simple assignment?
((int2=0))
((int1=0))

echo -n "please enter the first integer > "
read int1
echo -n "Please enter mathmatical operator > "
read input
echo -n "Please enter the second integer > "
read int2


if [ $input = $str1 ]
then
((int3=int1 + int2))
   echo "$int3"
else
   echo "sadly, it does not work"
fi

exec $SHELL

You should definitly checkout man bash. It is documented there in which command you need to specify $ or not to reference a variable. But aside from that:

var=123 # variable assignment. no spaces in between
echo $var # fetches/references the value of var. Or in other words $var gets substituted by it's value.

score 0 · Answer 2 · answered Feb 20 '17 at 16:58

0

Use bc command

something like this

echo "9/3+12" | bc

answered Feb 20 '17 at 16:58

nPcomp

8,637
2
54
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score 0 · Answer 3 · answered Feb 20 '17 at 16:58

0

You use $ when you want the value of the variable. read, though, expects the name of a variable:

read -p "..." int1

(Technically, you could do something like

name=int1
read -p "..." "$name"

to set the value of int1, because the shell expands name to the string int1, which read then uses as the name.)

answered Feb 20 '17 at 16:58

chepner

497,756
71
530
681

Good feedback, I have removed teh bash symbol from my variables, however I still get an error in execution – Anyon Feb 20 '17 at 17:02
@JonathanDeal: which error? What do you mean by the "bash symbol". The `$` should be considered to be a unary operator to give the value of a variable. – cdarke Feb 20 '17 at 17:08

score 0 · Answer 4 · answered Feb 20 '17 at 18:07

Here's a quick once over:

op=( add sub div mult )

int1=0
int2=0
ans=0

clear
read -p "please enter the first integer > " int1
clear
IFS='/'
read -p "Please enter mathmatical operator (${op[*]})> " input
unset IFS
clear
read -p "Please enter the second integer > " int2

case "$input" in
    add)  (( ans = int1 + int2 ));;
    sub)  (( ans = int1 - int2 ));;
    div)  (( ans = int1 / int2 ));; # returns truncated value and might want to check int2 != 0
    mult) (( ans = int1 * int2 ));;
    *)   echo "Invalid choice"
           exit 1;;
esac

echo "Answer is: $ans"

You will also want to check that the user enters numbers to :)

score 0 · Answer 5 · answered Feb 20 '17 at 18:36

Another one

declare -A oper
oper=([add]='+' [sub]='-' [div]='/' [mul]='*')

read -r -p 'Num1? > ' num1
read -r -p "oper? (${!oper[*]}) > " op
read -r -p 'Num2? > ' num2

[[ -n "${oper[$op]}" ]] || { echo "Err: unknown operation $op" >&2 ; exit 1; }
res=$(bc -l <<< "$num1 ${oper[$op]} $num2")
echo "$num1 ${oper[$op]} $num2 = $res"

How to perform arithmetic in UNIX?

5 Answers5