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| author | Sven Schnelle <svens@linux.ibm.com> | 2026-09-09 11:29:53 +0200 |
|---|---|---|
| committer | Masami Hiramatsu (Google) <mhiramat@kernel.org> | 2026-09-10 08:43:35 +0900 |
| commit | d22c3e0088e85be8131f7a9283f759cdbb20726d (patch) | |
| tree | 9668df2aa27b26f850d3ba5ba1287e443803675c /Documentation/ABI/removed/sysfs-kernel-uids | |
| download | linux-stable-d22c3e0088e85be8131f7a9283f759cdbb20726d.tar.gz linux-stable-d22c3e0088e85be8131f7a9283f759cdbb20726d.zip | |
selftests/ftrace: Fix unique symbol check in kprobe_non_uniq_symbol.tcgrafted
The current regex also matches symbols in modules, which makes the
test fail on s390 where name_show is present only once in the kernel,
but also multiple times in modules:
000001b1401cdc20 t name_show
000001b0c05e6c40 t name_show [mdev]
000001b0c0495f30 t name_show [i2c_core]
Fix this by changing the regular expression to only match the function
name.
Link: https://lore.kernel.org/all/20260909092954.2200558-1-svens@linux.ibm.com/
Fixes: 03b80ff8023a ("selftests/ftrace: Add new test case which checks non unique symbol")
Signed-off-by: Sven Schnelle <svens@linux.ibm.com>
Reviewed-by: Steven Rostedt <rostedt@goodmis.org>
Signed-off-by: Masami Hiramatsu (Google) <mhiramat@kernel.org>
Diffstat (limited to 'Documentation/ABI/removed/sysfs-kernel-uids')
| -rw-r--r-- | Documentation/ABI/removed/sysfs-kernel-uids | 14 |
1 files changed, 14 insertions, 0 deletions
diff --git a/Documentation/ABI/removed/sysfs-kernel-uids b/Documentation/ABI/removed/sysfs-kernel-uids new file mode 100644 index 000000000..85a90b86c --- /dev/null +++ b/Documentation/ABI/removed/sysfs-kernel-uids @@ -0,0 +1,14 @@ +What: /sys/kernel/uids/<uid>/cpu_shares +Date: December 2007, finally removed in kernel v2.6.34-rc1 +Contact: Dhaval Giani <dhaval@linux.vnet.ibm.com> + Srivatsa Vaddagiri <vatsa@linux.vnet.ibm.com> +Description: + The /sys/kernel/uids/<uid>/cpu_shares tunable is used + to set the cpu bandwidth a user is allowed. This is a + proportional value. What that means is that if there + are two users logged in, each with an equal number of + shares, then they will get equal CPU bandwidth. Another + example would be, if User A has shares = 1024 and user + B has shares = 2048, User B will get twice the CPU + bandwidth user A will. For more details refer + Documentation/scheduler/sched-design-CFS.rst |