What is -o in grep?

Shell Programming and Scripting
1

It was written in man grep:

-o
--only-matching

Print only the matched non-empty parts of matching lines, with each such part on a separate output line. Output lines use the same delimiters as input, and delimiters are null bytes if -z (--null-data) is also used (see [Other Options].

(a link in the manual removed so I could publish this post)

I didn't manage to understand this manual explanation.
Perhaps because English is not my first language.

What is the meaning of "matched non-empty part"?
Is it everything which is not a whitespace character or a tabloid character?

and by the way in my case I match just one line, not "lines".

Thanks

2

A demo text:

$ printf "%s\n" "hello world" "other stuff" "last line"
hello world
other stuff
last line

A standard grep prints the matchimg line:

$ printf "%s\n" "hello world" "other stuff" "last line" | grep "other"
other stuff

The -o prints only the matching part:

$ printf "%s\n" "hello world" "other stuff" "last line" | grep -o "other"
other

Here the matching part is just the given word.

The following prints the match of a regular expression:

$ printf "%s\n" "hello world" "other stuff" "last line" | grep -o "[aoui]the"
othe

Two matches in the line:

$ printf "%s\n" "hello world" "other stuff either" "last line" | grep -o "[aoui]the"
othe
ithe

It prints both matches, each on a separate output line.
While by default grep prints a matching line once:

$ printf "%s\n" "hello world" "other stuff either" "last line" | grep "[aoui]the"
other stuff either

I suggest you try out more examples yourself, until you have 100% clarity.

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