Received-SPF: pass (google.com: domain of linux-kernel-owner@vger.kernel.org designates 2620:137:e000::1:20 as permitted sender) client-ip=2620:137:e000::1:20;
MIME-Version: 1.0
References: <20230107034557.1156022-1-quanfafu@gmail.com> <30dd32b4-cd74-77a1-d9ad-84e361680dac@wanadoo.fr>
In-Reply-To: <30dd32b4-cd74-77a1-d9ad-84e361680dac@wanadoo.fr>
From:   Quanfa Fu <quanfafu@gmail.com>
Date:   Sat, 7 Jan 2023 20:23:57 +0800
Message-ID: <CAM4UoyqAGUbLaaoDyvFSTKvgpq9Oz7HuxKLMf3ryOsjHofcpVw@mail.gmail.com>
Subject: Re: [PATCH] tracing/eprobe: Replace kzalloc with kmalloc
To:     Christophe JAILLET <christophe.jaillet@wanadoo.fr>
Cc:     rostedt@goodmis.org, mhiramat@kernel.org,
        linux-kernel@vger.kernel.org, linux-trace-kernel@vger.kernel.org
Content-Type: text/plain; charset="UTF-8"
Content-Transfer-Encoding: quoted-printable
Precedence: bulk

On Sat, Jan 7, 2023 at 4:42 PM Christophe JAILLET
<christophe.jaillet@wanadoo.fr> wrote:
>
> Le 07/01/2023 =C3=A0 04:45, Quanfa Fu a =C3=A9crit :
> > Since this memory will be filled soon below, I feel that there is
> > no need for a memory of all zeros here. 'snprintf' does not return
> > negative num according to ISO C99, so I feel that no judgment is
> > needed here.
> >
> > No functional change intended.
> >
> > Signed-off-by: Quanfa Fu <quanfafu@gmail.com>
> > ---
> >   kernel/trace/trace_eprobe.c | 4 +---
> >   1 file changed, 1 insertion(+), 3 deletions(-)
> >
> > diff --git a/kernel/trace/trace_eprobe.c b/kernel/trace/trace_eprobe.c
> > index 352b65e2b910..cd1d271a74e7 100644
> > --- a/kernel/trace/trace_eprobe.c
> > +++ b/kernel/trace/trace_eprobe.c
> > @@ -917,15 +917,13 @@ static int trace_eprobe_parse_filter(struct trace=
_eprobe *ep, int argc, const ch
> >       for (i =3D 0; i < argc; i++)
> >               len +=3D strlen(argv[i]) + 1;
> >
> > -     ep->filter_str =3D kzalloc(len, GFP_KERNEL);
> > +     ep->filter_str =3D kmalloc(len, GFP_KERNEL);
> >       if (!ep->filter_str)
> >               return -ENOMEM;
> >
> >       p =3D ep->filter_str;
> >       for (i =3D 0; i < argc; i++) {
> >               ret =3D snprintf(p, len, "%s ", argv[i]);
> > -             if (ret < 0)
> > -                     goto error;
> >               if (ret > len) {
>
> Hi,
>
> as per [1]:
>   * The return value is the number of characters which would be
>   * generated for the given input, excluding the trailing null,
>   * as per ISO C99.  If the return is greater than *or equal* to
>   * @size, the resulting string is truncated.
>
> So, should this test be:
>      if (ret >=3D len)
>             ~~~~
>
>
> Also, isn't the p[-1] =3D '\0' after the loop eating the last character?
>     argc =3D 1;
>     argv[0] =3D "a";
>
>     Before the loop:
>     =3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D
>     len =3D 1 + 1 =3D 2;
>     ep->filter_str =3D 0x00 0x00
>                      ^
>                      |___ p
>
>     After the loop:
>     =3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D
>     ep->filter_str =3D 0x61 0x00
>                           ^
>                           |___ p
>     len =3D 1;
>
>     After p[-1]:
>     =3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D
>     ep->filter_str =3D 0x00 0x00
>                        ~~ ^
>                           |___ p
>
> Did I miss something obvious?
> I don't know the intent here, or if it is an issue at all, but it looks o=
dd.
>
> CJ
>
>
> [1]: https://elixir.bootlin.com/linux/v6.2-rc1/source/lib/vsprintf.c#L292=
5
> >                       ret =3D -E2BIG;
> >                       goto error;
>

I think that:

for example, argc =3D 2, argv =3D {"a", "b"};

Before the loop
=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D
len =3D (1 + 1) + (1 + 1) =3D 4;
ep->filter_str =3D 0x00 0x00 0x00 0x00
                       ^
                        |__ p

After the loop:
=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D
i =3D 1, snprintf write: 'a' and ' ', so ret =3D 2
i =3D 2, snprintf write: 'b' and ' ', so ret =3D 2
=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D
 Since the length of the last argv is not enough
to write, the space is replaced by null

ep->filter_str =3D 0x61 0x20 0x62 0x00
                                                          ^
                                                          |__ p
p =3D ep->fiter_str + 2 (ret1) +2 (ret2) =3D ep->filter_str + 4
=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D
so After p[-1] =3D *(ep->filter_str + 3) =3D '\0';
ep->filter_str =3D 0x61 0x20 0x62 0x00

According to ISO C99: " If the output was truncated due to this limit
then the return value is the number of characters (excluding the
terminating null byte) which would have been written to the final
string if enough  space  had been available"

The last snprintf will end with 'NULL', so I think p[-1] =3D '\0' can
also be deleted