[Linux] Differences between psutil.cpu_freq(True) and psutil.cpu_count()

[amanusk]

Rasbperry pi 3

• Raspbian
• latest master branch b23fb29

Bug description
Wonder about behaviour:

psutil.cpu_freq(percpu=True), on a Raspberry pi 3 now returns:
[scpufreq(current=600.0, min=600.0, max=1200.0)] i.e. only 1 CPU

When both psutil.cpu_count(logical=False) and psutil.cpu_count(logical=True) return 4: The expected behaviour.

Looking at the code, it seems this is because on the Pi,
there is only /sys/devices/system/cpu/cpufreq/policy0 while there are

/sys/devices/system/cpu/cpu0
/sys/devices/system/cpu/cpu1
/sys/devices/system/cpu/cpu2
/sys/devices/system/cpu/cpu3

And the former is checked first.
Should this be compared to psutil.cpu_count(logical=False), and the more accurate result taken?

Should I open a PR for this?

[giampaolo]

Mh... annoying. Looking back at how this code is implemented, it looks like we may want to try something different. Instead of glob we could use the CPU numbers directly. Something like this:

for x in range(psutil.cpu_count(logical=True)):
    if os.path.exists("/sys/devices/system/cpu/cpufreq/policy%s" % x):
        ...
    elif os.path.exists("/sys/devices/system/cpu/cpu%s/cpufreq" % x):
        ...
    else:
        break

It seems to me this will be more robust.

[giampaolo]

...or possibly this:

def cpu_freq():
    def get_path(num):
        for p in ("/sys/devices/system/cpu/cpufreq/policy%s" % num,
                  "/sys/devices/system/cpu/cpu%s/cpufreq" % num):
            if os.path.exists(p):
                return p

    for n in range(psutil.cpu_count(logical=True)):
        path = get_path(n)
        if not path:
            break
        ...