x86/pgtable: explain constant sign extension problem
When the _PAGE_FOO constants are defined as (1ul << _PAGE_BIT_FOO), they
become unsigned longs. In 32-bit PAE mode, these end up being
implicitly cast to 64-bit types when used to manipulate a pte, and
because they're unsigned the top 32-bits are 0, destroying the upper
bits of the pte.
When _PAGE_FOO constants are given a signed integer type, the cast to
64-bits will sign-extend so that the upper bits are all ones,
preserving the upper pte bits in manipulations.
Explain this in a prominent place.
Signed-off-by: Jeremy Fitzhardinge <jeremy@xensource.com>
Cc: Andi Kleen <ak@suse.de>
Signed-off-by: Ingo Molnar <mingo@elte.hu>
Signed-off-by: Thomas Gleixner <tglx@linutronix.de>
diff --git a/include/asm-x86/pgtable.h b/include/asm-x86/pgtable.h
index a1eeacd..75a656e 100644
--- a/include/asm-x86/pgtable.h
+++ b/include/asm-x86/pgtable.h
@@ -19,6 +19,11 @@
#define _PAGE_BIT_UNUSED3 11
#define _PAGE_BIT_NX 63 /* No execute: only valid after cpuid check */
+/*
+ * Note: we use _AC(1, L) instead of _AC(1, UL) so that we get a
+ * sign-extended value on 32-bit with all 1's in the upper word,
+ * which preserves the upper pte values on 64-bit ptes:
+ */
#define _PAGE_PRESENT (_AC(1, L)<<_PAGE_BIT_PRESENT)
#define _PAGE_RW (_AC(1, L)<<_PAGE_BIT_RW)
#define _PAGE_USER (_AC(1, L)<<_PAGE_BIT_USER)