Re: [PATCH v2 3/4] strbuf_setlen: don't write to strbuf_slopbuf

From: Brandon Casey <drafnel@gmail.com>
Date: Aug 23, 2017, 21:54 UTC
Message-ID: <CA+sFfMdMgrGBhECegBe09c38nRM+Zt5JK4gJaZ96DO-9zC-8qA@mail.gmail.com>
In-Reply-To: <CA+sFfMe56itAMDXOJybf0yHj+BqU1Ai1aU7inoTG3FJtdtZxyw@mail.gmail.com>

On Wed, Aug 23, 2017 at 2:20 PM, Brandon Casey <drafnel@gmail.com> wrote:

Show 13 quoted lines

> On Wed, Aug 23, 2017 at 2:04 PM, Junio C Hamano <gitster@pobox.com> wrote:
>> Brandon Casey <drafnel@gmail.com> writes:
>>
>>> So is there any reason why didn't do something like the following in
>>> the first place?
>>
>> My guess is that we didn't bother; if we cared, we would have used a
>> single instance of const char in a read-only segment, instead of
>> such a macro.
>
> I think you mean something like this:
>
>    const char * const strbuf_slopbuf = "";

Ah, you probably meant something like this:

   const char strbuf_slopbuf = '\0';

which gcc will apparently place in the read-only segment.  I did not know that.

And assignment and initialization would look like:

   sb->buf = (char*) &strbuf_slopbuf;

and

   #define STRBUF_INIT  { .alloc = 0, .len = 0, .buf = (char*) &strbuf_slopbuf }

respectively. Yeah, that's definitely preferable to a macro. Something similar could be done in object.c.

-Brandon