Does linux process VSZ equal 0 mean kernel space application?

Question

I notice some process always have VSZ as 0

USER       PID %CPU %MEM    VSZ   RSS TTY      STAT START   TIME COMMAND
root         1  0.0  0.0  19356  1400 ?        Ss   Jun13   0:00 /sbin/init
root         2  0.0  0.0      0     0 ?        S    Jun13   0:00 [kthreadd]
root         3  0.0  0.0      0     0 ?        S    Jun13   0:00 [migration/0]
root         4  0.0  0.0      0     0 ?        S    Jun13   0:01 [ksoftirqd/0]
root         5  0.0  0.0      0     0 ?        S    Jun13   0:00 [stopper/0]
root         6  0.0  0.0      0     0 ?        S    Jun13   0:03 [watchdog/0]
root         7  0.0  0.0      0     0 ?        S    Jun13   0:00 [migration/1]

how to understand why they have 0 VSZ?

Amit Verma · Accepted Answer · 2018-06-28T13:36:53.790

0

VSZ is the Virtual Memory Size. It includes all memory that the process can access, including memory that is swapped out, memory that is allocated, but not used, and memory that is from shared libraries.

So, the top command screenshot you shared showing VSZ values equaling 0, means that those processes are not using VSZ.

NOTE: They are kernel threads and memory statistics are irrelevant for them as they use kernel memory. Just to visualize kernel processes, press c when top command is running and it will show you all [bracketed] entries in last column named COMMAND.

You can get more details on VSZ and learn about its counterpart RSS (Resident Set Size) from here.

edited Jun 28 '18 at 13:36

answered Jun 26 '18 at 06:18

Amit Verma

8,660
8
35
40

for these processes not using VSZ, are they in kernel space NOT user space? – LinconFive Jun 27 '18 at 00:10
Yes, see more in the note section, I have added in my answer above. – Amit Verma Jun 28 '18 at 13:37

Does linux process VSZ equal 0 mean kernel space application?

1 Answers1