enh: assume CPU is immune to Zenbleed regardless of vendor except AMD

This contradicts our usual "if we don't know, consider vulnerable" motto,
but as this vuln is extremely specific (which is not the case for the Spectre
range of vulnerabilities, for example), this is the correct approach here.
This commit is contained in:
Stéphane Lesimple 2023-07-26 17:52:33 +02:00
parent c0869d7341
commit f34dd5fa7b
1 changed files with 2 additions and 5 deletions

View File

@ -352,7 +352,8 @@ is_cpu_affected()
variant_taa=''
variant_itlbmh=''
variant_srbds=''
variant_zenbleed=''
# Zenbleed if extremely AMD specific, look for "is_and" below:
variant_zenbleed=immune
if is_cpu_mds_free; then
[ -z "$variant_msbds" ] && variant_msbds=immune
@ -385,7 +386,6 @@ is_cpu_affected()
variant_mdsum=immune
variant_taa=immune
variant_srbds=immune
variant_zenbleed=immune
elif is_intel; then
# Intel
# https://github.com/crozone/SpectrePoC/issues/1 ^F E5200 => spectre 2 not affected
@ -461,7 +461,6 @@ is_cpu_affected()
_debug "is_cpu_affected: intel family < 6 is immune to l1tf"
[ -z "$variantl1tf" ] && variantl1tf=immune
fi
variant_zenbleed=immune
elif is_amd || is_hygon; then
# AMD revised their statement about variant2 => affected
# https://www.amd.com/en/corporate/speculative-execution
@ -478,7 +477,6 @@ is_cpu_affected()
variantl1tf=immune
# Zenbleed
variant_zenbleed=immune
amd_legacy_erratum "$(amd_model_range 0x17 0x30 0x0 0x4f 0xf)" && variant_zenbleed=vuln
amd_legacy_erratum "$(amd_model_range 0x17 0x60 0x0 0x7f 0xf)" && variant_zenbleed=vuln
amd_legacy_erratum "$(amd_model_range 0x17 0xa0 0x0 0xaf 0xf)" && variant_zenbleed=vuln
@ -587,7 +585,6 @@ is_cpu_affected()
_debug "is_cpu_affected: for cpu$i and so far, we have <$variant1> <$variant2> <$variant3> <$variant3a> <$variant4>"
done
variantl1tf=immune
variant_zenbleed=immune
fi
# we handle iTLB Multihit here (not linked to is_specex_free)