Improve the utility function unload_module
In some cases, a simple 'modprobe -r' won't remove a kernel module with
dependencies. This patch makes module removal more robust, handling
cases that simple 'modprobe -r' won't cut it.
Impact: Low (users of the function unload_module will notice it handles
better removal of modules with dependencies).
Signed-off-by: Mike Burns <[email protected]>
git-svn-id: http://test.kernel.org/svn/autotest/trunk@3219 592f7852-d20e-0410-864c-8624ca9c26a4
diff --git a/client/bin/base_utils.py b/client/bin/base_utils.py
index 139a094..d6099dc 100755
--- a/client/bin/base_utils.py
+++ b/client/bin/base_utils.py
@@ -8,7 +8,7 @@
precedence order defined there
"""
import os, shutil, sys, signal, commands, pickle, glob, statvfs
-import math, re, string, fnmatch
+import math, re, string, fnmatch, logging
from autotest_lib.client.common_lib import error, utils
@@ -567,7 +567,24 @@
def unload_module(module_name):
- utils.system('/sbin/rmmod ' + module_name)
+ """
+ Removes a module. Handles dependencies. If even then it's not possible
+ to remove one of the modules, it will trhow an error.CmdError exception.
+
+ @param module_name: Name of the module we want to remove.
+ """
+ l_raw = utils.system_output("/sbin/lsmod").splitlines()
+ lsmod = [x for x in l_raw if x.split()[0] == module_name]
+ if len(lsmod) > 0:
+ line_parts = lsmod[0].split()
+ if len(line_parts) == 4:
+ submodules = line_parts[3].split(",")
+ for submodule in submodules:
+ unload_module(submodule)
+ utils.system("/sbin/modprobe -r %s" % module_name)
+ logging.info("Module %s unloaded" % module_name)
+ else:
+ logging.info("Module %s is already unloaded" % module_name)
def module_is_loaded(module_name):
