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

[Wacek Kusnierczyk]

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.

vQ