How can I add numbers in a Bash script?

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`

Use the $(( )) arithmetic expansion.

num=$(( $num + $metab ))

See Chapter 13. Arithmetic Expansion for more information.

There are a thousand and one ways to do it. Here's one using dc (a reverse Polish desk calculator which supports unlimited precision arithmetic):

dc <<<"$num1 $num2 + p"

But if that's too bash-y for you (or portability matters) you could say

echo $num1 $num2 + p | dc

But maybe you're one of those people who thinks RPN is icky and weird; don't worry! bc is here for you:

bc <<< "$num1 + $num2"
echo $num1 + $num2 | bc

That said, there are some unrelated improvements you could be making to your script:

#!/bin/bash

num=0
metab=0

for ((i=1; i<=2; i++)); do
    for j in output-$i-* ; do # 'for' can glob directly, no need to ls
            echo "$j"

             # 'grep' can read files, no need to use 'cat'
            metab=$(grep EndBuffer "$j" | awk '{sum+=$2} END { print sum/120}')
            num=$(( $num + $metab ))
    done
    echo "$num"
done

As described in Bash FAQ 022, Bash does not natively support floating point numbers. If you need to sum floating point numbers the use of an external tool (like bc or dc) is required.

In this case the solution would be

num=$(dc <<<"$num $metab + p")

To add accumulate possibly-floating-point numbers into num.

In Bash,

 num=5
 x=6
 (( num += x ))
 echo $num   # ==> 11

Note that Bash can only handle integer arithmetic, so if your AWK command returns a fraction, then you'll want to redesign: here's your code rewritten a bit to do all math in AWK.

num=0
for ((i=1; i<=2; i++)); do
    for j in output-$i-*; do
        echo "$j"
        num=$(
           awk -v n="$num" '
               /EndBuffer/ {sum += $2}
               END {print n + (sum/120)}
           ' "$j"
        )
    done
    echo "$num"
done

I always forget the syntax so I come to Google Search, but then I never find the one I'm familiar with :P. This is the cleanest to me and more true to what I'd expect in other languages.

i=0
((i++))

echo $i;