sed DON'T remove extra whitespace - bash

Question

It seems everybody else wants to remove any additional whitespace, however I have the opposite problem.
I have a file, call it some_file.txt that looks like
a b c d
and some more
and I'm reading it line-by-line with sed,
num_lines=$(cat some_file.txt | wc -l)
for i in $(seq 1 $num_lines); do
echo $(sed "${i}q;d" $file)
string=$(sed "${i}q;d" $file)
echo $string
done
I would expect the number of whitespace characters to stay the same, however the output I get is
a b c d
a b c d
and some more
and some more
So it seems that the problem is with sed removing the extra whitespace between chars, anyway to fix this?

Answer 1

Have a look at this example:
$ echo Hello World
Hello World
$ echo "Hello World"
Hello World
sed is not your problem, your problem is that bash removes the whitespaces when passing the output of sed into echo.
You just need to surround whatever echo is supposed to print with double quotation marks. So instead of
echo $(sed "${i}q;d" $file)
echo $string
You write
echo "$(sed "${i}q;d" $file)"
echo "$string"
The new script should look like this:
#!/usr/bin/env bash
file=some_file.txt
num_lines=$(cat some_file.txt | wc -l)
for i in $(seq 1 $num_lines); do
echo "$(sed "${i}q;d" $file)"
string=$(sed "${i}q;d" $file)
echo "$string"
done
prints the correct output:
a b c d
a b c d
and some more
and some more
However, if you just want to go through your file line by line, I strongly recommend something like this:
while IFS= read -r line; do
echo "$line"
done < some_file.txt
Question from the comments: What to do if you only want 33 lines starting from line x. One possible solution is this:
#!/usr/bin/env bash
declare -i s=$1
declare -i e=${s}+32
sed -n "${s},${e}p" $file | while IFS= read -r line; do
echo "$line"
done
(Note that I would probably include some validation of $1 in there as well.)
I declare s and e as integer variables, then even bash can do some simple arithmetic on them and calculate the actual last line to print.