[PATCH] vi: fix regex search compilation error

[dietmar.schindler at manrolandgoss.com]

> -----Original Message-----
> From: busybox <busybox-bounces at busybox.net> On Behalf Of Bernhard Reutner-Fischer
> Sent: Wednesday, July 14, 2021 12:51 AM
>
> On Tue, 13 Jul 2021 21:42:05 +0100
> Ron Yorston <rmy at pobox.com> wrote:
>
> > Bernhard Reutner-Fischer wrote:
> > >If i'm not mistaken, the s++ is fully redundant here although it seems
> > >that my gcc-11 does not optimize it like if we manually spell it out
> > >like in the attached?
> >
> > I don't think it's possible to avoid two increments.  When a backslash
> > is detected we want to move past it and the following character.
> >
> > Consider:
> >
> >    strchr_backslash("\\/abc/def", '/');
> >
> > The original code and Denys' rewrite return a pointer to "/def"; your
> > suggestion returns a pointer to "/abc/def".
>
> @@ -2686,10 +2686,9 @@ static char *strchr_backslash(const char *s, int
> c) while (*s) {
>  if (*s == c)
>  return (char *)s;
> -if (*s == '\\') # 1
> -if (*++s == '\0') # 2
> +if (*s++ == '\\') # 3
> +if (*s == '\0') # 4
>  break;
> -s++; # 5
>  }
>  return NULL;
>
> How come? Obviously i'm slow today and in addition blind.
> So..
> 1) if *s == backslash
> increment s
> old: if the incremented s points to 0 (when we break, #2)
> new: always (#3)
> 2) If *s was a backslash
> old: if the incremented *s was 0 (#2)
> new: if the previously incremented *s was 0 (#3)
> then break and return NULL
>
> ah oh in the old code we double increment (only) if the first was a
> backslash _not_ followed by a nul.
> hmz. So is there a more elegant way to express that which i don't see
> right now?

I can't say whether that's more elegant, only that it's equivalent:

while (*s)
{
if (*s == c)
return (char *)s;
if (*s != '\\' || *++s)
s++;
}
 return NULL;
--
Best regards,
Dietmar Schindler
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