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authorDavid Hildenbrand <david@redhat.com>2019-05-31 16:33:38 +0200
committerDavid Hildenbrand <david@redhat.com>2019-06-07 14:53:25 +0200
commite19a61eb514dbf7c9a725c7539ce3b6166cd6ac4 (patch)
treea5e38ff7b5e094ed51db00d79e064afae964402b /target/s390x/mmu_helper.c
parent88a29e867528c57151103e64a966a9cbaeec852f (diff)
s390x/tcg: Store only the necessary amount of doublewords for STFLE
The PoP (z14, 7-382) says: Doublewords to the right of the doubleword in which the highest-numbered facility bit is assigned for a model may or may not be stored. However, stack protection in certain binaries can't deal with that. "gzip" example code: f1b4: a7 08 00 03 lhi %r0,3 f1b8: b2 b0 f0 a0 stfle 160(%r15) f1bc: e3 20 f0 b2 00 90 llgc %r2,178(%r15) f1c2: c0 2b 00 00 00 01 nilf %r2,1 f1c8: b2 4f 00 10 ear %r1,%a0 f1cc: b9 14 00 22 lgfr %r2,%r2 f1d0: eb 11 00 20 00 0d sllg %r1,%r1,32 f1d6: b2 4f 00 11 ear %r1,%a1 f1da: d5 07 f0 b8 10 28 clc 184(8,%r15),40(%r1) f1e0: a7 74 00 06 jne f1ec <file_read@@Base+0x1bc> f1e4: eb ef f1 30 00 04 lmg %r14,%r15,304(%r15) f1ea: 07 fe br %r14 f1ec: c0 e5 ff ff 9d 6e brasl %r14,2cc8 <__stack_chk_fail@plt> In QEMU, we currently have: max_bytes = 24 the code asks for (3 + 1) doublewords == 32 bytes. If we write 32 bytes instead of only 24, and return "2 + 1" doublewords ("one less than the number of doulewords needed to contain all of the facility bits"), the example code detects a stack corruption. In my opinion, the code is wrong. However, it seems to work fine on real machines. So let's limit storing to the minimum of the requested and the maximum doublewords. Cc: Stefan Liebler <stli@linux.ibm.com> Cc: Andreas Krebbel <Andreas.Krebbel@de.ibm.com> Reviewed-by: Richard Henderson <richard.henderson@linaro.org> Signed-off-by: David Hildenbrand <david@redhat.com>
Diffstat (limited to 'target/s390x/mmu_helper.c')
0 files changed, 0 insertions, 0 deletions