1. Oracle VM Server for SPARC 3.6 Administration Guide
  2. Performing Other Administration Tasks
  3. Determining Where Errors Occur by Mapping CPU and Memory Addresses
  4. Example of CPU and Memory Mapping

Example of CPU and Memory Mapping

Example 18-4 Determining the Configuration of Domains

The following command produces a long parseable list of logical domains configurations.

primary# ldm list -l -p
VERSION 1.6
DOMAIN|name=primary|state=active|flags=normal,control,vio-service|
cons=SP|ncpu=4|mem=1073741824|util=0.6|uptime=64801|
softstate=Solaris running
VCPU
|vid=0|pid=0|util=0.9|strand=100
|vid=1|pid=1|util=0.5|strand=100
|vid=2|pid=2|util=0.6|strand=100
|vid=3|pid=3|util=0.6|strand=100
MEMORY
|ra=0x8000000|pa=0x8000000|size=1073741824
IO
|dev=pci@780|alias=bus_a
|dev=pci@7c0|alias=bus_b
...
DOMAIN|name=ldg1|state=active|flags=normal|cons=5000|
ncpu=2|mem=805306368|util=29|uptime=903|
softstate=Solaris running
VCPU
|vid=0|pid=4|util=29|strand=100
|vid=1|pid=5|util=29|strand=100
MEMORY
|ra=0x8000000|pa=0x48000000|size=805306368
...
DOMAIN|name=ldg2|state=active|flags=normal|cons=5001|
ncpu=3|mem=1073741824|util=35|uptime=775|
softstate=Solaris running
VCPU
|vid=0|pid=6|util=35|strand=100
|vid=1|pid=7|util=34|strand=100
|vid=2|pid=8|util=35|strand=100
MEMORY
|ra=0x8000000|pa=0x78000000|size=1073741824
...

Example 18-5 Determining the Virtual CPU That Corresponds to a Physical CPU Number

The logical domain configuration is shown in Example 18-4. This example describes how to determine the domain and the virtual CPU corresponding to physical CPU number 5, and the domain and the real address corresponding to physical address 0x7e816000.

Looking through the VCPU entries in the list for the one with the pid field equal to 5, you can find the following entry in the stanza for logical domain ldg1. 

|vid=1|pid=5|util=29|strand=100

Hence, the physical CPU number 5 is in domain ldg1 and within the domain it has virtual CPU number 1.

Looking through the MEMORY entries in the list, you can find the following entry in the stanza for domain ldg2. 

ra=0x8000000|pa=0x78000000|size=1073741824

Where 0x78000000 <= 0x7e816000 <= (0x78000000 + 1073741824 - 1); that is, pa <= PA <= (pa + size - 1). Hence, the PA is in domain ldg2 and the corresponding real address is 0x8000000 + (0x7e816000 - 0x78000000) = 0xe816000.