On Wednesday 03 October 2007 16:18, H. Peter Anvin wrote: > 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.
How does the compiler know it doesn't depend on memory? How do you say it depends on memory? You really need something as heavy as volatile? - To unsubscribe from this list: send the line "unsubscribe linux-kernel" in the body of a message to [EMAIL PROTECTED] More majordomo info at http://vger.kernel.org/majordomo-info.html Please read the FAQ at http://www.tux.org/lkml/
