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Date: Fri, 07 Sep 2018 18:00:46 +0200
From: Steffen Nurpmeso <steffen@...oden.eu>
To: Rich Felker <dalias@...c.org>
Cc: musl@...ts.openwall.com
Subject: Re: Regex: behaviour of ? after () atom
Rich Felker wrote in <20180907153302.GM1878@...ghtrain.aerifal.cx>:
|On Fri, Sep 07, 2018 at 05:25:17PM +0200, Steffen Nurpmeso wrote:
|> Rich Felker wrote in <20180907151821.GL1878@...ghtrain.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:]]+)(.*)$> 17
|>|> i=<print 1 2> -> <> <> <>
|>|>
|>|> It works correctly if i remove the ()? atom, so i thought i should
|>|> report that.
|>|
|>|What is the value of the flags argument you passed to regcomp?
|>|
|>
|> REG_EXTENDED, optional REG_ICASE:
|>
|> reflrv = REG_EXTENDED;
|> if(f & a_ICASE)
|> reflrv |= REG_ICASE;
|> if((reflrv = regcomp(&re, argv[2], reflrv))){
|
|OK, it looks like that should work, and seemed to work here when I
|passed the regex to grep -E linked with musl's regex. Can you provide
|a minimal self-contained C program to demonstrate the issue you're
|having?
Happy user that i am, here something for tests/:
#include <stdio.h>
#include <regex.h>
int main(void){
regmatch_t rema[1 + 21];
regex_t re;
int i;
i = REG_EXTENDED;
if((i = regcomp(&re, "^(:[[:space:]]+)?([^[:space:]]+)(.*)$", i)))
return 2;
i = regexec(&re, "print 1 2", 21, rema, 0);
regfree(&re);
if(i == REG_NOMATCH)
return 3;
for(i = 1; i < 21 && rema[i].rm_so != -1; ++i)
;
return (i == 3) ? 0 : 4;
}
i is 1 here.
|BTW which "()?" are you talking about? The whole first parenthesized
|subsexpression and the ? after it? I wouldn't call that an atom, but
|nothing seems wrong with it.
I have read regex(7) first just in case something intellectual had
to be said. Otherwise i am all for Finnish tango.
--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)
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