integer arithmetic - How can I add numbers in a Bash script?

Question

I have this Bash script and I had a problem in line 16. How can I take the previous result of line 15 and add it to the variable in line 16?

#!/bin/bash

num=0
metab=0

for ((i=1; i<=2; i++)); do
    for j in `ls output-$i-*`; do
        echo "$j"

        metab=$(cat $j|grep EndBuffer|awk '{sum+=$2} END { print sum/120}') (line15)
        num= $num + $metab   (line16)
    done
    echo "$num"
 done

Answer 1

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)

  num= 1 num =2

• bc and expr expect each number and operator as a separate argument, so whitespace is important. They cannot process arguments like 3+ +4.

  num=`expr $num1+ $num2`