On Tue, Mar 29, 2011 at 10:06, Junio C Hamano <gitster@xxxxxxxxx> wrote:
> The fetch-pack/upload-pack protocol relies on the underlying transport
> (local pipe or TCP socket) to have enough slack to allow one window worth
> of data in flight without blocking the writer. Traditionally we always
> relied on being able to have a batch of 32 "have"s in flight (roughly 1.5k
Its 64. Because the client "races ahead" one window before ever
reading. Which is closer to 3K of data in flight.
> The recent "progressive-stride" change allows "fetch-pack" to send up to
> 1024 "have"s without reading any response from "upload-pack". The
> outgoing pipe of "upload-pack" can be clogged with many ACK and NAK that
> are unread, while "fetch-pack" is still stuffing its outgoing pike with
s/pike/pipe/
> @@ -229,16 +229,17 @@ static void insert_alternate_refs(void)
> }
>
> #define INITIAL_FLUSH 16
> +#define PIPESAFE_FLUSH 32
> #define LARGE_FLUSH 1024
>
> static int next_flush(int count)
> {
> - if (count < INITIAL_FLUSH * 2)
> - count += INITIAL_FLUSH;
> - else if (count < LARGE_FLUSH)
> + int flush_limit = args.stateless_rpc ? LARGE_FLUSH : PIPESAFE_FLUSH;
> +
> + if (count < flush_limit)
> count <<= 1;
> else
> - count += LARGE_FLUSH;
> + count += flush_limit;
Nak. You still deadlock because when count reaches PIPESAFE_FLUSH you
still double it to 2*PIPESAFE_FLUSH here. Instead I think you mean:
if (args.stateless_rpc) {
if (count < LARGE_FLUSH)
count <<= 1;
else
count += LARGE_FLUSH;
} else {
if (count * 2 < PIPESAFE_FLUSH)
count <<= 1;
}
--
Shawn.
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