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authorMartin Schwidefsky <schwidefsky@de.ibm.com>2017-04-04 12:55:11 +0200
committerMartin Schwidefsky <schwidefsky@de.ibm.com>2017-04-05 10:11:38 +0200
commitdf26c2e87e6cf3ced1fbd589e40d633a6a7f20cb (patch)
tree296d3e12a690cdb9c91dbe639caf0530a5811497 /arch/s390/include/asm/spinlock.h
parents390/pci: remove forward declaration (diff)
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s390/cpumf: simplify detection of guest samples
There are three different code levels in regard to the identification of guest samples. They differ in the way the LPP instruction is used. 1) Old kernels without the LPP instruction. The guest program parameter is always zero. 2) Newer kernels load the process pid into the program parameter with LPP. The guest program parameter is non-zero if the guest executes in a process != idle. 3) The latest kernels load ((1UL << 31) | pid) with LPP to make the value non-zero even for the idle task. The guest program parameter is non-zero if the guest is running. All kernels load the process pid to CR4 on context switch. The CPU sampling code uses the value in CR4 to decide between guest and host samples in case the guest program parameter is zero. The three cases: 1) CR4==pid, gpp==0 2) CR4==pid, gpp==pid 3) CR4==pid, gpp==((1UL << 31) | pid) The load-control instruction to load the pid into CR4 is expensive and the goal is to remove it. To distinguish the host CR4 from the guest pid for the idle process the maximum value 0xffff for the PASN is used. This adds a fourth case for a guest OS with an updated kernel: 4) CR4==0xffff, gpp=((1UL << 31) | pid) The host kernel will have CR4==0xffff and will use (gpp!=0 || CR4!==0xffff) to identify guest samples. This works nicely with all 4 cases, the only possible issue would be a guest with an old kernel (gpp==0) and a process pid of 0xffff. Well, don't do that.. Suggested-by: Christian Borntraeger <borntraeger@de.ibm.com> Reviewed-by: Christian Borntraeger <borntraeger@de.ibm.com> Signed-off-by: Martin Schwidefsky <schwidefsky@de.ibm.com>
Diffstat (limited to 'arch/s390/include/asm/spinlock.h')
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