Adding adding and listing elements of an array together in Bash

Question

I am very new to bash and am currently stuck on a small project. I am trying to list the combined scores of each array (num1 and num2) and then display the output using a C-style loop. Here is my desired output:

Result_1: 71
Result_2: 96
Result_3: 101
Result_4: 86
Result_5: 148

As you can see, the first number in the num1 array is added with the first number of the num2 array to form Result_1, and the second number in the num1 array is added with the second number from the num2 array to form Result_2.. and so on. However, I am only able to get and display the sum of the first two numbers (13 and 58 = 71) for all results when this is not what I am after. I also wish to have the first result as "Result_1" instead of "Result_0", but can't seem to figure this out either. This is my current code:

#!/bin/bash

num1=(13 28 11 72 50)
num2=(58 68 90 14 98)
len=${#num1[*]}

for (( i=0; i<${len}; i++ ));
do
    sum=$(($num1 + $num2))
    echo "Result_"$i":" $sum
done

Any help would be greatly appreciated. Thanks!

Taylor G. · Accepted Answer · 2021-04-23 04:55:43Z

3

You just forgot to link index to array. The following loop will do the correct job.

for (( i=0; i<${len}; i++ ))
do
    sum=$((${num1[$i]} + ${num2[$i]}))
    echo "Result_$(($i+1)):" $sum
done

answered Apr 23, 2021 at 4:55

Taylor G.

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