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

[Wacek Kusnierczyk]

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'.  there is no 'line
> before the match';  the line before that containing '1' is *within* the
> match.
>   

to follow up, that's what you'd see in perl:

    "a\n1\n" =~ /^(?:|.*[^0])1/

evaluates to 1 (true). 

but on reflection, i'd agree there is a sort-of problem, in that man
grep says (emphasis added):

"    grep  searches the named input FILEs (or standard input if no files are
       named, or if a single hyphen-minus (-) is given as file name)
*for lines*
       containing  a  match to the given PATTERN.  By default, grep
prints the
       matching lines."

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.

vQ