Re: [Qemu-devel] Memory mapping on MIPS

[Igor R]

> > Here is an excerpt from r4k_map_address(), related to addresses >= 0x80000000.
> > Actually, it maps 0x80000010 and 0xA0000010 to the same physical
> > address. What's the idea behind that?
>
> 0x80000010 is kseg0 whereas 0xA0000010 is kseg1, both segments are
> unmapped thus both refer to the same 0x10 physical address. The
> difference between them is that the latter is uncached, i.e. all
> accesses to this segment bypass the caches, but in QEMU we don't model
> caches so they do the same thing.
>
> > What should happen if I map KSEG2 directly as a continuation of KSEG1,
> > i.e. substitute TLB lookup with "address - (int32_t)KSEG1_BASE"? Guest
> > Linux seems to work correctly (but maybe it's just a matter of luck?).
>
> With such a change it's very likely that guest through kseg2 will be
> accessing different physical addresses than it expects...

I see, thanks.
As far as I can see now, my problem is that cpu_memory_rw_debug doesn't handle tlb misses at all - that's why it fails for many addresses in kseg2.
Is it the desired behavior or a bug? Is there any other simple way to read from the guest virt.addrs?

Thanks!