Using the $(( )) arithmetic expansion.
num=$(( $num + $metab ))
For integers:
Use arithmetic expansion: $((EXPR)) ```
num=$((num1 + num2))
num=$(($num1 + $num2)) # Also works
num=$((num1 + 2 + 3)) # ...
num=$[num1+num2] # Old, deprecated arithmetic expression syntax ```
Using the external expr utility. Note that this is only needed for really old systems. ```
num=`expr $num1 + $num2` # Whitespace for expr is important ``` For floating point:
Bash doesn’t directly support this, but there are a couple of external tools you can use:
num=$(awk "BEGIN {print $num1+$num2; exit}")
num=$(python -c "print $num1+$num2")
num=$(perl -e "print $num1+$num2")
num=$(echo $num1 + $num2 | bc) # Whitespace for echo is important
You can also use scientific notation (for example, 2.5e+2).
Common pitfalls:
When setting a variable, you cannot have whitespace on either side of =, otherwise it will force the shell to interpret the first word as the name of the application to run (for example, num= or num)
bc and expr expect each number and operator as a separate argument, so whitespace is important. They cannot process arguments like 3+ +4.
A=1
B=1
let "C = $A + $B"
echo $C # C == 2
echo $num1 $num2 + p | dc
dc <<<"$num1 $num2 + p"
num=$(dc <<<"$num $metab + p")