Linux bash. for loop and function, for adding numbers
Try this:
n=$1
sum=0
for i in `seq 1 $n` ; do
## redefine variable 'sum' after each iteration of for-loop
sum=`expr $sum + $i`
done
echo $sum
With a while loop and similar to your code:
#!/bin/bash
n=$(expr $1 + 1)
result=0
j=0
add(){
result=$(expr $result + $j)
}
while test $j -ne $n
do
add
j=$(expr $j + 1)
done
echo $result
The $(..whatever..) is similar to `..whatever..`, it executes your command and returns the value. The test command is very usefull, check out the man. In this case simulates a for loop comparing the condition $j -ne $n (j not equal n) and adding 1 to the j var in each turn of the loop.
You could try below:
#!/usr/bin/env bash
sumit() {
local n=$1
local sum=0
for (( i=0;i<=n;i++ )) ; do
(( sum = sum + i ))
done
echo "$sum"
}
sum=$(sumit $1)
echo "sum is ($sum)"