Nick Piggin wrote:
This should work because the result gets used before reading again:
read_cr3(a);
write_cr3(a | 1);
read_cr3(a);
But this might be reordered so that b gets read before the write:
read_cr3(a);
write_cr3(a | 1);
read_cr3(b);
?
I don't see how, as write_cr3 clobbers memory.
Because read_cr3() doesn't depend on memory, and b could be stored in a
register.
-hpa
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