How to perform arithmetic in UNIX? - bash

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;

Answer 1

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.

Answer 2

Use bc command
something like this
echo "9/3+12" | bc

Answer 3

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.)

Answer 4

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 :)

Answer 5

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"