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From: Steffen Nurpmeso <steffen@sdaoden.eu>
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Subject: Re: Regex: behaviour of ? after () atom
Date: Fri, 07 Sep 2018 17:25:17 +0200
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Rich Felker wrote in <20180907151821.GL1878@brightrain.aerifal.cx>:
 |On Fri, Sep 07, 2018 at 03:38:05PM +0200, Steffen Nurpmeso wrote:
 |> Hello.
 |> 
 |> In perl this is
 |> 
 |>   $x="print 1 2";
 |>   if($x =~ /^(:[[:space:]]+)?([^[:space:]]+)(.*)$/){ 
 |>     print "<$0> -> <$1> <$2> <$3>\n"
 |>}
 |> 
 |> and the result is
 |> 
 |>   </tmp/t.pl> -> <> <print> < 1 2>
 |> 
 |> Now the same on AlpineLinux edge and musl-1.1.19-r10 with the MUA
 |> i maintain, which uses the normal regex stuff and calls it via
 |> 
 |>   echo eins=$3
 |>          vput vexpr i regex "${3}" \
 |>             '^(:[[:space:]]+)?([^[:space:]]+)(.*)$'  \
 |>             '<\$0> -> <\$1> <\$2> <\$3>'
 |>   echo i=$i
 |> 
 |> which in C code does 
 |> 
 |>       if((reflrv = regcomp(&re, argv[2], reflrv))){
 |>           ...
 |>          goto jestr;
 |>}
 |>   fprintf(stderr, "GOING for <%s> -> <%s> %u\n",
 |>   argv[1],argv[2],n_NELEM(rema));
 |>       reflrv = regexec(&re, argv[1], n_NELEM(rema), rema, 0);
 |> 
 |> and overall prints
 |> 
 |>   eins=print 1 2
 |>   GOING for <print 1 2> -> <^(:[[:space:]]+)?([^[:space:]]+)(.*)$> 17
 |>   i=<print 1 2> -> <> <> <>
 |> 
 |> It works correctly if i remove the ()? atom, so i thought i should
 |> report that.
 |
 |What is the value of the flags argument you passed to regcomp?
 |

REG_EXTENDED, optional REG_ICASE:

      reflrv = REG_EXTENDED;
      if(f & a_ICASE)
         reflrv |= REG_ICASE;
      if((reflrv = regcomp(&re, argv[2], reflrv))){


--steffen
|
|Der Kragenbaer,                The moon bear,
|der holt sich munter           he cheerfully and one by one
|einen nach dem anderen runter  wa.ks himself off
|(By Robert Gernhardt)