From: Stafford Horne <shorne@gmail.com>
To: Richard Henderson <richard.henderson@linaro.org>
Cc: QEMU Development <qemu-devel@nongnu.org>,
"Jason A . Donenfeld" <Jason@zx2c4.com>
Subject: Re: [PATCH] target/openrisc: Do not reset delay slot flag on early tb exit
Date: Thu, 12 May 2022 07:34:51 +0900 [thread overview]
Message-ID: <Ynw6CymBBiqQgElu@antec> (raw)
In-Reply-To: <9c8c68d0-cb7c-4ec5-eac4-fd915934db5e@linaro.org>
On Wed, May 11, 2022 at 02:56:37PM -0700, Richard Henderson wrote:
> On 5/11/22 14:43, Stafford Horne wrote:
> > At this point how would tb->flags have the right value? Would it always be set
> > correctly by `cpu_get_tb_cpu_state` in the `lookup_tb_ptr`call?
>
> Well, it would be set by cpu_get_tb_cpu_state in cpu_exec, which is then
> passed to tb_gen_code. If we go around a loop and look it up a second time,
> we'll find a tb with a matching set of flags.
Right, cpu_get_tb_cpu_state called in lookup_tb_ptr will not update tb->flags.
What you mention, that is for when we have to generate a new TB, the tb->flags
get set right before tb_gen_code.
But for the case where we exit the delay-slot TB due to a pending exception I
think the flow would go.
TB chain:
-> branch-TB : set env->flag 1
-> delay-slot-TB : exit_tb due to condition
Exit:
-> return to cpu_tb_exec
-> tcg_ops->synchronize_from_tb
In this case I don't see how the tb->flag would be updated, ooh, I guess it
would have been set earlier when the TB was generated. Maybe that is what I am
missing.
-Stafford
next prev parent reply other threads:[~2022-05-11 22:36 UTC|newest]
Thread overview: 7+ messages / expand[flat|nested] mbox.gz Atom feed top
2022-05-11 12:05 [PATCH] target/openrisc: Do not reset delay slot flag on early tb exit Stafford Horne
2022-05-11 14:32 ` Richard Henderson
2022-05-11 21:43 ` Stafford Horne
2022-05-11 21:56 ` Richard Henderson
2022-05-11 22:34 ` Stafford Horne [this message]
2022-05-12 2:11 ` Richard Henderson
2022-05-12 12:47 ` Stafford Horne
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