How to ensure SysRq is always enabled regardless of the kernel.sysrq setting?

Question

Some Linux distributions have kernel.sysrq=16 which means only SysRq + s (sync) is allowed. As an example: Fedora (25 and 28) has it set as such in /usr/lib/sysctl.d/50-default.conf

I had to create a file as /etc/sysctl.d/95-sysrq.conf where I manually set kernel.sysrq=1 so it's available as soon as possible (but possibly not soon enough depending on situation):

$ grep -nHi sysrq /usr/lib/sysctl.d/* /etc/sysctl.d/*
/usr/lib/sysctl.d/50-default.conf:16:# Use kernel.sysrq = 1 to allow all keys.
/usr/lib/sysctl.d/50-default.conf:17:# See http://fedoraproject.org/wiki/QA/Sysrq for a list of values and keys.
/usr/lib/sysctl.d/50-default.conf:18:kernel.sysrq = 16
/etc/sysctl.d/95-sysrq.conf:1:kernel.sysrq=1

Is there a way to enable Sysrq from early boot, possibly also ignoring any setting for kernel.sysrq?, for example adding a kernel boot parameter (e.g., cat /proc/cmdline for current ones) such as from the Grub boot menu (or in xen.cfg's kernel= line).

Answer 1

There is kernel boot parameter sysrq_always_enabled according to the doc:

sysrq_always_enabled
            [KNL]
            Ignore sysrq setting - this boot parameter will
            neutralize any effect of /proc/sys/kernel/sysrq.
            Useful for debugging.

I have tested Sysrq to work(eg. help,sync) even when kernel.sysctl=0 (so it's just as the doc above says), if I add kernel boot argument sysrq_always_enabled (note: it doesn't have to be sysrq_always_enabled=1).
(to double check, I've also tested Sysrq to NOT work when kernel.sysctl=0 and sysrq_always_enabled is NOT present in /proc/cmdline)
Source code confirms it too.

Note: /proc/sys/kernel/sysrq is another way to read/write kernel.sysrq

For more info on sysrq: https://www.kernel.org/doc/html/v4.15/admin-guide/sysrq.html
(though sysrq_always_enabled is not mentioned there, but it is here)

EDIT: When sysrq_always_enabled is in effect, there is a dmesg line:

[    0.000000] sysrq: sysrq always enabled.