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cups: fix CVE-2022-26691
In scheduler/cert.c the previous algorithm didn't expect the strings can have a different length, so one string can be a substring of the other and such substring was reported as equal to the longer string. Backport patch from upstream to fix: https://github.com/OpenPrinting/cups/commit/de4f8c196106033e4c372dce3e91b9d42b0b9444 CVE: CVE-2022-26691 (From OE-Core rev: cc657868d31cc8b4218a07aa10fa098c379e473c) Signed-off-by: Steve Sakoman <steve@sakoman.com> Signed-off-by: Richard Purdie <richard.purdie@linuxfoundation.org>
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committed by
Richard Purdie
parent
238fb89434
commit
95cda9d091
@@ -13,6 +13,7 @@ SRC_URI = "https://github.com/apple/cups/releases/download/v${PV}/${BP}-source.t
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file://0002-don-t-try-to-run-generated-binaries.patch \
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file://0003-cups_1.4.6.bb-Fix-build-on-ppc64.patch \
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file://0004-cups-fix-multilib-install-file-conflicts.patch\
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file://CVE-2022-26691.patch \
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"
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UPSTREAM_CHECK_URI = "https://github.com/apple/cups/releases"
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@@ -119,4 +120,4 @@ cups_sysroot_preprocess () {
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# -25317 concerns /var/log/cups having lp ownership. Our /var/log/cups is
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# root:root, so this doesn't apply.
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CVE_CHECK_WHITELIST += "CVE-2021-25317"
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CVE_CHECK_WHITELIST += "CVE-2021-25317"
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@@ -0,0 +1,33 @@
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From de4f8c196106033e4c372dce3e91b9d42b0b9444 Mon Sep 17 00:00:00 2001
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From: Zdenek Dohnal <zdohnal@redhat.com>
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Date: Thu, 26 May 2022 06:27:04 +0200
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Subject: [PATCH] scheduler/cert.c: Fix string comparison (fixes
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CVE-2022-26691)
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The previous algorithm didn't expect the strings can have a different
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length, so one string can be a substring of the other and such substring
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was reported as equal to the longer string.
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CVE: CVE-2022-26691
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Upstream-Status: Backport [https://github.com/OpenPrinting/cups/commit/de4f8c196106033e4c372dce3e91b9d42b0b9444]
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Signed-off-by: Steve Sakoman
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---
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diff --git a/scheduler/cert.c b/scheduler/cert.c
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index b268bf1b2..9b65b96c9 100644
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--- a/scheduler/cert.c
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+++ b/scheduler/cert.c
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@@ -434,5 +434,12 @@ ctcompare(const char *a, /* I - First string */
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b ++;
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}
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- return (result);
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+ /*
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+ * The while loop finishes when *a == '\0' or *b == '\0'
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+ * so after the while loop either both *a and *b == '\0',
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+ * or one points inside a string, so when we apply logical OR on *a,
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+ * *b and result, we get a non-zero return value if the compared strings don't match.
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+ */
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+
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+ return (result | *a | *b);
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}
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