Messages in this thread | | | Date | Mon, 28 Mar 2005 10:22:07 -0800 | From | "H. J. Lu" <> | Subject | Re: i386/x86_64 segment register issuses (Re: PATCH: Fix x86 segment register access) |
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On Mon, Mar 28, 2005 at 09:46:00AM -0800, H. J. Lu wrote: > On Mon, Mar 28, 2005 at 05:47:06PM +0200, Andi Kleen wrote: > > "H. J. Lu" <hjl@lucon.org> writes: > > > The new assembler will disallow them since those instructions with > > > memory operand will only use the first 16bits. If the memory operand > > > is 16bit, you won't see any problems. But if the memory destinatin > > > is 32bit, the upper 16bits may have random values. The new assembler > > > > Does it really have random values on existing x86 hardware? > > The x86 hardwares will only change the first 16bits. The rest bits > are unchanged. A simple test program can verify that. > > > > > If it is a only a "theoretical" problem that does not happen > > in practice I would advise to not do the change. > > > > It depends on what the initial value in the upper bits is. The > assembler in CVS generates the same binary code as > > movw %ds,(%eax) > > for > > movl %ds,(%eax) > > But the previous assemblers will generate > > 66 8c 18 movw %ds,(%eax) > > for > > movw %ds,(%eax) > > This bug has been fixed for a while. I guess that may be why Linux > kernel uses > > movl %ds,(%eax)
It turns out that both old and new assemblers will generate
0: 8c 18 movw %ds,(%eax)
for mov %ds,(%eax)
So kernel can use "mov" instead of "movl" and the binary output will be the same.
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