Python script to determine if Linux is running on a virtual machine

In the Webhostingtalk forum, some foreign businessmen will be virtual machine as a standalone server to sell, last July, when a Chinese fellow was deceived, and in WHT posted denounced, evidence, even the service providers themselves also admitted that the replies up to 355 articles. This standalone server/vps provider hostatree.com incredibly bold OpenVZ VPS such a look on the virtual machine as a standalone server sold, dizzy, at least to get a vmware/kvm/xen HVM it (more difficult to find is a virtual machine), with OpenVZ This kind of container is also too bullying: Yesterday, just received a netizen e-mail asked how to determine whether they bought a standalone server or virtual machine problems. Here Vpsee a brief introduction to the market on the common virtual technology (including container technology) discrimination tips.

Judging Openvz/xen pv/uml

Judge Openvz/xen Pv/uml is the easiest, directly check the relevant directories and files under/proc can be known, such as OpenVZ VPS will have/proc/vz this file, Xen PV virtual machine will have/proc/xen/this directory, and the directory has a The UML logo is found on the UML print/proc/cpuinfo. Wrote a simple Python script to detect:

The code is as follows:

#!/usr/bin/python
# Check if a Linux system running on a virtual machine (Openvz/xen pv/uml)

Import sys, OS

def main ():
If Os.getuid ()! = 0:
Print "must be run as root"
Sys.exit (0)

# Check Openvz/virtuozzo
If Os.path.exists ("/proc/vz"):
If not os.path.exists ("/PROC/BC"):
Print "OpenVZ container"
Else
Print "OpenVZ node"

# Check Xen
If Os.path.exists ("/proc/xen/capabilities"):
if (Os.path.getsize ("/proc/xen/capabilities") > 0):
Print "Xen dom0"
Else
Print "Xen DomU"

# check User Mode Linux (UML)
f = open ("/proc/cpuinfo", "R"); t = F.read (); F.close ()
if (T.find ("UML") > 0):
Print "UML"

If __name__== "__main__":
Main ()

Judging Vmware/xen HVM/KVM

If the use of Vmware/xen HVM/KVM such a full virtual is more difficult to judge some, the most accurate way is to read the CPUID to determine, the Xen source code below is a detection is Xen C code TOOLS/MISC/XEN-DETECT.C, this code provides A good example of this, Vpsee rewritten the code, replaced the function with a macro, added the recognition of VMware and KVM, and then compiled it with GCC to run:

The code is as follows:

/*
* Check if a Linux system running on a virtual machine (Vmware/xen HVM/KVM)
*/
#include stdio.h
#include string.h

#define HYPERVISOR_INFO 0x40000000

#define CPUID (idx, eax, ebx, ecx, edx) \
ASM volatile (\
"Test%1,%1; JZ 1f; ud2a;. ASCII \ "Xen\"; 1:cpuid "\
: "=b" (*EBX), "=a" (*eax), "=c" (*ECX), "=d" (*edx) \
: "0" (idx));

int main (void)
{
unsigned int eax, EBX, ecx, edx;
Char string[13];

CPUID (Hypervisor_info, &eax, &EBX, &ECX, &edx);
* (unsigned int *) (string+0) = EBX;
* (unsigned int *) (string+4) = ECX;
* (unsigned int *) (string+8) = edx;

STRING[12] = 0;
if (strncmp (String, "xenvmmxenvmm", 12) = = 0) {
printf ("Xen hvm\n");
} else if (strncmp (String, "Vmwarevmware", 12) = = 0) {
printf ("vmware\n");
} else if (strncmp (String, "KVMKVMKVM", 12) = = 0) {
printf ("kvm\n");
} else
printf ("Bare hardware\n");

return 0;
}

Judging virtualbox/virtual PC

What the? This kind of Home desktop virtual machine installed itself will not know?! If you do not know, there is a way to run the Dmidecode tool under Linux and then find Manufacturer:innotek GmbH, the manufacturer:microsoft Corporation keyword can correspond to Virtualb Ox and Virtual PC.

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