bug fix for softmmu slow_st unaligned access
PATCH 1/1 bug fix for softmmu slow_st unaligned access
For example, the memory address 0x10008000 is on an unwritable page.When the instruction "add dword ptr [0x10007FFF],0x12345678" is executed,the OS will set 0x10008000 page be a writable page and re-execute this instruction. But softmmu has modifed the value of 0x10007FFF,so after re-executing this instruction, the final result is wrong(double-added on 0x10007FFF). Reversing the stored byte order can fix this bug.
softmmu.patch:
*** qemu.orig/softmmu_template.h Sun Oct 28 11:15:52 2007
--- qemu/softmmu_template.h Sun Oct 28 11:22:24 2007
*************** static void glue(glue(slow_st, SUFFIX), *** 282,288 ****
} else if (((addr & ~TARGET_PAGE_MASK) + DATA_SIZE - 1) >=
TARGET_PAGE_SIZE) {
do_unaligned_access:
/* XXX: not efficient, but simple */! for(i = 0;i < DATA_SIZE; i++) {
#ifdef TARGET_WORDS_BIGENDIAN
glue(slow_stb, MMUSUFFIX)(addr + i, val >> (((DATA_SIZE -
1) * 8) - (i * 8)),
mmu_idx, retaddr);
--- 282,288 ----
} else if (((addr & ~TARGET_PAGE_MASK) + DATA_SIZE - 1) >=
TARGET_PAGE_SIZE) {
do_unaligned_access:
/* XXX: not efficient, but simple */! for(i = DATA_SIZE-1;i >= 0; i--) {
#ifdef TARGET_WORDS_BIGENDIAN
glue(slow_stb, MMUSUFFIX)(addr + i, val >> (((DATA_SIZE -
1) * 8) - (i * 8)),
mmu_idx, retaddr);
http://www.nabble.com/file/p13449885/softmmu.patch softmmu.patch
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Hi,
Quoted 6 lines · TeLeMan
For example, the memory address
0x10008000is on an unwritable page.When the instruction "add dword ptr [0x10007FFF],0x12345678" is executed,the OS will set0x10008000page be a writable page and re-execute this instruction. But softmmu has modifed the value of 0x10007FFF,so after re-executing this instruction, the final result is wrong(double-added on0x10007FFF). Reversing the stored byte order can fix this bug.
I'm not sure I understand, but what happens if now the 10008000 page is writable and 10007fff isn't, thus the OS needs to make it writable and re-execute? I guess reversing the accesses order is not a solution?
Regards,
Andrew
Quoted 14 lines · andrzej zaborowski
Hi,
Quoted 9 lines · TeLeMan
For example, the memory address
0x10008000is on an unwritable page.When the
instruction "add dword ptr [0x10007FFF],0x12345678" is executed,the OS will
set0x10008000page be a writable page and re-execute this instruction. But
softmmu has modifed the value of 0x10007FFF,so after re-executing this instruction, the final result is wrong(double-added on0x10007FFF). Reversing the stored byte order can fix this bug.
Quoted 9 lines
I'm not sure I understand, but what happens if now the 10008000 page is writable and
10007fffisn't, thus the OS needs to make it writable and re-execute? I guess reversing the accesses order is not a solution?Regards,
Andrew
If the 0x10008000 page is writable and 0x10007FFF isn't, softmmu can raise this exception before modifing 0x10007FFF-0x10008002 because softmmu checks 0x10007FFF at first. I don't know if reversing the order is an exact solution,but its simple and working.
btw, I found this bug because I found the some windows dll reloc offset are calculated incorrectly by the guest OS. If you need a sample, I can give it to you.
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Hi,
Quoted 4 lines · TeLeMan
If the
0x10008000page is writable and0x10007FFFisn't, softmmu can raise this exception before modifing 0x10007FFF-0x10008002 because softmmu checks0x10007FFFat first. I don't know if reversing the order is an exact solution,but its simple and working.
Assuming that DATA_SIZE is less than page size, I think it is okay for a workaround, I committed it. It shows that permissions are checked two times for the first byte (0x10007ffff) but maybe that's okay since this is a slow access anyway.
Regards