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`

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Use the $(( )) arithmetic expansion.

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

See Chapter 13. Arithmetic Expansion for more information.

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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.

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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

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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;

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I really like this method as well. There is less clutter:

count=$[count+1]

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 #!/bin/bash
read X
read Y
echo "$(($X+$Y))"

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You should declare metab as integer and then use arithmetic evaluation

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

For more information, see 6.5 Shell Arithmetic.

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Use the shell built-in let. It is similar to (( expr )):

A=1
B=1
let "C = $A + $B"
echo $C # C == 2

Source: Bash let builtin command

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Another portable POSIX compliant way to do in Bash, which can be defined as a function in .bashrc for all the arithmetic operators of convenience.

addNumbers () {
    local IFS='+'
    printf "%s\n" "$(( $* ))"
}

and just call it in command-line as,

addNumbers 1 2 3 4 5 100
115

The idea is to use the Input-Field-Separator(IFS), a special variable in Bash used for word splitting after expansion and to split lines into words. The function changes the value locally to use word-splitting character as the sum operator +.

Remember the IFS is changed locally and does not take effect on the default IFS behaviour outside the function scope. An excerpt from the man bash page,

The shell treats each character of IFS as a delimiter, and splits the results of the other expansions into words on these characters. If IFS is unset, or its value is exactly , the default, then sequences of , , and at the beginning and end of the results of the previous expansions are ignored, and any sequence of IFS characters not at the beginning or end serves to delimit words.

The "$(( $* ))" represents the list of arguments passed to be split by + and later the sum value is output using the printf function. The function can be extended to add scope for other arithmetic operations also.

---

#!/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

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#!/usr/bin/bash

#integer numbers
#===============#

num1=30
num2=5

echo $(( num1 + num2 ))
echo $(( num1-num2 ))
echo $(( num1*num2 ))
echo $(( num1/num2 ))
echo $(( num1%num2 ))

read -p "Enter first number : " a
read -p "Enter second number : " b
# we can store the result
result=$(( a+b ))
echo sum of $a \& $b is $result # \ is used to espace &


#decimal numbers
#bash only support integers so we have to delegate to a tool such as bc
#==============#

num2=3.4
num1=534.3

echo $num1+$num2 | bc
echo $num1-$num2 | bc
echo $num1*$num2 |bc
echo "scale=20;$num1/$num2" | bc
echo $num1%$num2 | bc

# we can store the result
#result=$( ( echo $num1+$num2 ) | bc )
result=$( echo $num1+$num2 | bc )
echo result is $result

##Bonus##
#Calling built in methods of bc 

num=27

echo "scale=2;sqrt($num)" | bc -l # bc provides support for calculating square root

echo "scale=2;$num^3" | bc -l # calculate power

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