In the boot decompressor for the kernel in the image Iouri provided, I
32bit or 64bit image?
As you can plainly see, the call to memcpy (which is redefined in
boot/compressed/misc.c) is made using stack calling convention.
Unfortunately, the compiler generated the memcpy function itself using
regparm(3) convention. I am guessing this happened because a leftover
I can't find any here grepping .i files and includes. None of the memcpy
prototypes in include have a __fastcall
Are you sure this still happens in mainline?
When I look at my scroll it also looks correct:
c0: 0f af f8 imul %eax,%edi
c3: 01 c0 add %eax,%eax
c5: 89 c2 mov %eax,%edx
c7: 01 f2 add %esi,%edx
c9: 89 45 f0 mov %eax,0xfffffff0(%ebp)
cc: 89 f0 mov %esi,%eax
ce: 89 f9 mov %edi,%ecx
d0: e8 7b ff ff ff call 50 <memcpy>
Does anyone recall compiler problems (I noticed this VM was booting a
64-bit kernel,
Ok 64bit, that was 32bit above.
Since some time the decompressor is 64bit and 64bit always uses regparms. 64bit Scroll is ok too:
92: 0f af eb imul %ebx,%ebp
95: 01 db add %ebx,%ebx
97: 48 63 f3 movslq %ebx,%rsi
9a: 4c 01 ee add %r13,%rsi
9d: 89 ea mov %ebp,%edx
9f: e8 9c ff ff ff callq 40 <memcpy>
but the decompressor was 32-bit code,
That must be a old kernel.
Can you people please verify this all still happens with a mainline or
2.6.22 kernel?
so perhaps at some time the 64-bit makefile or toolchain for Linux had some kind of
compiler bug).
I'm not aware of any such bugs.