How can I add numbers in a Bash script?

user_798353 0
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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
mathematical-expressions integer-arithmetic bash
12 Answers
 #!/bin/bash
read X
read Y
echo "$(($X+$Y))"
user_68818390

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

user_40868710 
#!/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
user_82348700
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