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Re: [Qemu-devel] The reason behind block linking constraint?
From: |
Max Filippov |
Subject: |
Re: [Qemu-devel] The reason behind block linking constraint? |
Date: |
Mon, 26 Sep 2011 01:47:36 +0400 |
User-agent: |
KMail/1.13.7 (Linux/2.6.40.4-5.fc15.x86_64; KDE/4.6.5; x86_64; ; ) |
> > I meant TLB change by e.g. tlb_set_page. If you change single page
> > mapping then all TBs in that page will be gone.
> > This may be the result of e.g. a page swapping, or a task switch.
>
> You said "all TBs in that page will be gone". Does it mean QEMU will
> invalidate those TBs by for example, tb_invalidate_phys_page_range? Or
> it just leave those TBs there (without executing them)? Because I don't
> see functions like tb_invalidate_phys_page_range are called after
> tlb_set_page, I am not sure what "gone" actually means.
Well, my explanation sucks. Let's say it other way, more precisely:
- you have two pieces of code in different pages, one of them jumps to the
other;
- and you have two TBs, tb1 for the first piece and tb2 for the second;
- and you link them and there's direct jump from tb1 to tb2;
- now you change the mapping of the code page that contains second piece of
code;
- after that there's another code (or no code at all) at the place where the
second piece of code used to be;
- but the jump to tb2 still remains in tb1.
> > If there's no direct link between TBs then softmmu will be used during
> > the target TB search and softmmu will generate an appropriate guest
> > exception. See cpu_exec -> tb_find_fast -> tb_find_slow ->
> > get_page_addr_code.
> >
> > But if there is a direct link, then softmmu has no chance to do it.
>
> Let me try to describe the flow. Correct me if I am wrong. Assume tb1
> and tb2 belong to different guest pages.
>
> If there's NO direct link between tb1 and tb2. After executing tb1,
> the control is transfered back to QEMU (cpu_exec), QEMU then call
> tb_find_fast to find the next TB, i.e., tb2.
Right.
> I assume that "all TBs in that page will be gone" means QEMU will
> invalidate those TBs.
No, it won't. I had to say "all code in that page will be gone", sorry for the
confusion.
> If not, I think tb_find_fast will return tb2 which should not be executed.
It won't either. tb_find_fast searches tb this way:
tb = env->tb_jmp_cache[tb_jmp_cache_hash_func(pc)];
but 'page mapping change' implies TLB flush, at least for that page.
Both tlb_flush and tlb_flush_page will clear env->tb_jmp_cache and tb_find_fast
will have to call tb_find_slow.
> So, tb_find_fast calls tb_find_slow,
> then tb_find_slow calls get_page_addr_code.
>
> get_page_addr_code return the guest physical address which
> corresponds to the pc (tb2). It looks up TLB (env1->tlb_table), and
> get a TLB miss since tlb_set_page has changed the mapping.
>
> if (unlikely(env1->tlb_table[mmu_idx][page_index].addr_code !=
> (addr & TARGET_PAGE_MASK))) {
> ldub_code(addr);
> }
>
> But I am not sure what happen after the TLB miss. You said the
> softmmu will generate a guest exception. Take x86 as an example,
> do you mean the raise_exception_err in tlb_fill?
>
> void tlb_fill(target_ulong addr, ...) {
> ret = cpu_x86_handle_mmu_fault(env, addr, is_write, mmu_idx, 1);
> if (ret) {
> if (retaddr) {
> }
> }
> raise_exception_err(env->exception_index, env->error_code);
> }
Exactly. The exception will be raised inside the guest and the guest will
execute its page fault handler or whatever.
Thanks.
-- Max