| To: | Aurelien Jarno <aurelien@aurel32.net> |
|---|---|
| Subject: | Re: [Qemu-devel] QEMU/MIPS & dyntick kernel |
| From: | Alan Cox <alan@lxorguk.ukuu.org.uk> |
| Date: | Tue, 2 Oct 2007 21:48:26 +0100 |
| Cc: | qemu-devel@nongnu.org, linux-mips@linux-mips.org |
| In-reply-to: | <4702AC0F.1000906@aurel32.net> |
| Organization: | Red Hat UK Cyf., Amberley Place, 107-111 Peascod Street, Windsor, Berkshire, SL4 1TE, Y Deyrnas Gyfunol. Cofrestrwyd yng Nghymru a Lloegr o'r rhif cofrestru 3798903 |
| Original-recipient: | rfc822;linux-mips@linux-mips.org |
| References: | <20071002200644.GA19140@hall.aurel32.net> <4702A99B.7020008@qumranet.com> <4702AC0F.1000906@aurel32.net> |
| Sender: | linux-mips-bounce@linux-mips.org |
> Well on real hardware, the instruction rate and the timer are linked: > the timer run at half the speed of the CPU. As the corresponding > assembly code is very small, only uses registers and is run in kernel > mode, you know for sure that 48 cycles is more than enough. What happens on NMI or if you take an ECC exception and scrubbing stall off the memory controller while loading part of that cache line of code into memory ? Alan |
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