William <william.duc...@ensimag.grenoble-inp.fr> writes:
> On Mon, May 30, 2016 at 11:34:42PM -0700, Junio C Hamano wrote:
>
>> As long as your "on stack strbuf" allows lengthening the string
>> beyond the initial allocation (i.e. .alloc, not .len), the user of
>> the API (i.e. the one that placed the strbuf on its stack) would not
>> know when the implementation (i.e. the code in this patch) decides
>> to switch to allocated memory, so it must call strbuf_release()
>> before it leaves. Which in turn means that your implementation of
>> strbuf_release() must be prepared to be take a strbuf that still has
>> its string on the stack.
>
> Well, my implementation does handle a strbuf that still has its
> string on the stack: the buffer won't be freed in this case (only a
> reset to STRBUF_INIT).
> Unless I misunderstood you?
I think Junio meant:
void f()
{
struct strbuf sb;
char buf[N];
strbuf_wrap_preallocated(&sb, buf, ...);
strbuf_add(&sb, ...);
// is it safe to just let sb reach the end of its scope?
}
To answer the last question, the user would need to know too much about
the allocation policy, so in this case the user should call
strbuf_release(), and let it chose whether to call "free()"
(OWNS_MEMORY) or not. This is OK with your implementation, but the doc
needs to reflect this.
--
Matthieu Moy
http://www-verimag.imag.fr/~moy/
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