for loop in bash exiting to early without errors - bash

Question

Why doesn't my for loop: for ((i=1 ;i<=$n, i++)) work? I can not figure this out. when n is for example 4, the line echo $n returns 4 but it does not go into the loop again. I don't get any errors either. I tried to make a small loop like:
for ((i=1, i<=$n; i++)); do
echo "this works"
done
This works fine which makes it even stranger to me :/. Thanks in advance
read n
length=16
p=()
p[1]=50
rest=63
function s() {
arr=($#)
line="_____________________________________________________________________"
for i in ${arr[#]}; do
line=$( echo $line | sed s/./1/$i)
done
echo $line
}
for ((i=1; i<=$n; i++)); do
echo $n
for ((j=1; j<=$length; j++)); do
s ${p[#]}
done
len=${#p[#]}
((len=$len*2))
for ((k=1; k<=$len; k+=2)); do
((p[$k+1]=p[$k]+1))
((p[$k]=p[$k]-1))
done
for ((l=1; l<=$length; l++)); do
s ${p[#]}
len=${#p[#]}
for ((m=1; m<=$len; m+=2)); do
((p[$m+1]=p[$m+1]+1))
((p[$m]=p[$m]-1))
done
done
((rest=$rest-2*$length))
((length=$length/2))
done

Answer 1

If you don't declare your variables local, they're global -- so the loop in your s function is overwriting the same i counter used outside the function, leading to the outer loop's early exit.
Consider using the below code:
s() {
local -a arr # declare a function-local array (sparse, like all bash arrays)
local i # ...and a function-local counter
for i; do # by default, a for statement iterates over "$#"; that works for us.
arr[$i]=1 # specifically set a value only for named items
done
# ...thereafter, iterate through the range of characters we want to print...
for ((i=0; i<70; i++)); do
printf '%s' "${arr[$i]:-_}" # and print the array entry if present, or an _ otherwise
done
printf '\n' # ...followed by a trailing newline.
}