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

There are a thousand and one ways to do it. Here's one using dc:

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 a 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

EDIT:

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.

Answer 2

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's 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 (e.g.: 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 (eg: 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`

Answer 3

#!/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+=$2} END { print sum/120}') (line15)
        let num=num+metab (line 16)
    done
    echo "$num"
done

Answer 4

You should declare metab as integer and then use arithmetic evaluation

declare -i metab num
...
num+=metab
...

For more information see https://www.gnu.org/software/bash/manual/html_node/Shell-Arithmetic.html#Shell-Arithmetic