Re: [bug #25765] grep -P regexp also prints the line before the match

[Dave B]

Wacek Kusnierczyk wrote:

> Wacek Kusnierczyk wrote:
>> Andriy Sen wrote:
>>   
>>> See tree examples below. The last one shows the problem.
>>>
>>>
>>> G:\>cat test.s
>>> a
>>> 1
>>>
>>> G:\>od -a test.s
>>> 0000000   a  nl   1  nl
>>> 0000004
>>>
>>> G:\>cat test.s | grep -P "^[^0]1"
>>>
>>>
>>> G:\>cat test.s | grep -P "^(|[^0])1"
>>> 1
>>>
>>> G:\>cat test.s | grep -P "^(|.*[^0])1"
>>> a
>>> 1
>>>
>>>   
>>>     
>> what problem? 
>>
>>     cat test.s | grep -Pn "^(|.*[^0])1"
>>     # 1:a
>>     # 1
>>
>> as you can see, there is one match here.  '.*' matches 'a', and '[^0]'
>> matches the newline.  and so '1' can match '1'.

Uhm, grep is supposed to not cross line boundaries, even with -P I'd say...

Using Perl the result is different:

$ printf 'a\n1\n' | perl -ne 'print if /^(|.*[^0])1/'
1
$

So I'd expect grep -P to produce the same result...which it doesn't.

(and, besides that, there's the issue of an empty alternation being used,
whose results might not always be specified, see eg POSIX)

> clearly, your test.s does not contain any single *line* that would match
> the pattern -- but that's a problem with the documentation, i'd think,
> which hasn't been updated to accommodate for the -P option which can
> cause grep to match *across* lines, as above, and as here:
> 
>     grep '.\n.' test.s
>     # no match
>     grep -P '.\n.' test.s
>     # match
> 
> i would certainly not consider this a bug, though.

Perl does not match across lines by default, you have to use the /s flag in
the match operator. Anyway, you might be correct, but then why doesn't grep
-P show that behavior consistently?

$ printf 'a\n1\n' | grep -P 'a.1'
$

$ printf 'a\n1\n' | grep -P 'a.*1'
$

If grep -P matched across lines, I'd expect the last two examples to output

a
1

and instead they don't output anything.

-- 
D.