Explain please sed script

Question

I can't understand how this sed script works:

echo -e "Line #1\n\n\n\nLine #2\n\n\nLine #3" | sed '1s/^$//p;/./,/^$/!d'

It suppresses repeated empty lines like cat -s But I have a couple of questions:

For what 1s/^$//p? As I understand it do nothing with the first line even if it empty
Is this /./,/^$/ matches only before first ^$ like Line #1\n\n and not matches Line #1\n\n\n?
Are ranges not greedy by default in sed?

To clarify question 3 I tried next tests:

echo -e "Line #1\n\n\n\nLine #2\n\n\nLine #3" | sed -n '/#/,/#/p'

And result was:

Line #1 Line #2 Line #3

(so, it is greedy)

But when I tried:

echo -e "Line #1\n\n\n\nLine #2\n\n\nLine #3" | sed -n '/#1/,/#/p'

result was:

Line #1 Line #2

(now it seems to be not greedy)

@don_crissti, it's not nonsense, it checks if the line is empty and prints it if so. — ilkkachu
– ilkkachu, Commented Feb 20, 2018 at 22:49
@don_crissti, in GNU sed, you could use 1{/^$/p}, but some other seds (like the one on my Mac) may be more picky about requiring newlines with the braces. — ilkkachu
– ilkkachu, Commented Feb 20, 2018 at 22:58

ilkkachu · Accepted Answer · 2018-02-20 22:55:43Z

1s/^$//p prints the first line, if it's empty.

/./,/^$/ matches lines from the first non-empty line, to the first empty line encountered. It's not greedy in the sense that a regex qualifier is: sed can't look ahead to the file or backtrack, so it has to stop the first time the ending pattern matches.

After the ending line, the search for the beginning line starts again, so the next non-empty line again starts the range. In effect, the range matches contiguous nonempty lines, plus the first following empty one.

Since the range is used as /./,/^$/!d, all lines not matching it are deleted. This includes the very first line if it's empty, which is why it's explicitly printed by the first rule.

Without the 1s/^$//p rule, the first line would be removed if empty, even though it's not really "repeating".

$ echo $'\nfoo' | sed '1s/^$//p;/./,/^$/!d' foo $ echo $'\nfoo' | sed '/./,/^$/!d' foo $

In your test, the range /#/,/#/ is a bit different since it starts and ends with the same pattern. Line #1 matches the beginning pattern, (so the intervening empty lines are printed) Line #2 matches the ending one, (the following empty lines aren't) and on Line #3, the range begins again.

In the other one, the starting pattern is /#1/, but that's only found once in the input.

Paul T · Accepted Answer · 2018-02-21 03:39:29Z

ilkkachu gave a great answer, which covered every aspect of this command, including the less-than-obvious need for the first part of the sed string; and including the fact that the "!" character deletes the parts that DON'T match, in practice the repeating blank lines.

Still, it's an odd way to do it, when there are easier options, like more -s, or uniq available.

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