Re: "grep -o" skips some matching patterns

[Roberto Gordo Saez]

On Thu, Jan 01, 2004 at 01:04:26PM +0100, Stepan Kasal wrote:
> Hello,
>       thank very much to Michael for the explanation, which helped.
> 
> I'd like to add a theoretical comment which won't help at all ;-)
> 
> On Tue, Dec 30, 2003 at 04:24:23PM +0100, Roberto Gordo Saez wrote:
> > $ echo abc0111def | grep -o "[01]*"
> 
> This command should in fact print seven lines, three empty ones,
> then the non-empty match and another three empty ones.
> The reason can be explained by the following (the sed command performs
> global substitution of your regex by "(...)"):
> 
> $ echo abc0111def|sed 's/[01]*/(&)/g'
> ()a()b()c(0111)d()e()f()
> 
> So you have actually triggered a bug in ``grep -o''.
> 
> I hope to help with fixing it...
> 
> Stepan Kasal

So it is true, it has a bug :-)

grep stops with the first empty match, because the non-empy
ones at the begining are printed. It is easy to find why (from grep.c):

|  if (only_matching)
|    {
|      size_t match_size;
|      size_t match_offset;
|      while ((match_offset = (*execute) (beg, lim - beg, &match_size, 1))
|          != (size_t) -1)
|        {
|          char const *b = beg + match_offset;
|          if (b == lim)
|            break;
|          if (match_size == 0)
|            break;
           ^^^^^^^^^^^^^^^^^^^^
Here, match_size is zero. As a test, doing { beg = b + 1; continue; }
instead of "break" will print all non-empty matchs (That is not a fix,
it is only a test, i know nothing about grep source code).

|          if(color_option)
|            printf("\33[%sm", grep_color);
|          fwrite(b, sizeof (char), match_size, stdout);
|          if(color_option)
|            fputs("\33[00m", stdout);
|          fputs("\n", stdout);
|          beg = b + match_size;
|        }

-- 
Roberto Gordo Saez - Free Software Engineer
Linalco "Especialistas en Linux y Software Libre"
http://www.linalco.com/  Tel: +34-914561700