Use heuristics to run sepolicy_tests faster
We are compiling regex more than 300000 times, and it's a main
bottleneck for slow sepolicy_tests. Actually we don't need to compile
regex that much; most of cases can be handled by simple string
comparison. This change introduces heuristics for optimization.
Bug: 301874100
Test: verified that return values of MatchPathPrefix are not changed.
Test: run cProfile, before and after.
Before
ncalls tottime percall cumtime percall filename:lineno(function)
21951 0.923 0.000 56.491 0.003 policy.py:33(MatchPathPrefix)
After
ncalls tottime percall cumtime percall filename:lineno(function)
21951 0.078 0.000 1.159 0.000 policy.py:40(MatchPathPrefix)
Change-Id: I1ebad586c2518e74a8ca67024df5e77d068e3ca5
diff --git a/tests/policy.py b/tests/policy.py
index 805c451..9fdc43c 100644
--- a/tests/policy.py
+++ b/tests/policy.py
@@ -30,7 +30,46 @@
# 1) there is a match - return True or 2) run out of characters - return
# False.
#
+COMMON_PREFIXES = {
+ "/(vendor|system/vendor)": ["/vendor", "/system/vendor"],
+ "/(odm|vendor/odm)": ["/odm", "/vendor/odm"],
+ "/(product|system/product)": ["/product", "/system/product"],
+ "/(system_ext|system/system_ext)": ["/system_ext", "/system/system_ext"],
+}
+
def MatchPathPrefix(pathregex, prefix):
+ # Before running regex compile loop, try two heuristics, because compiling
+ # regex is too expensive. These two can handle more than 90% out of all
+ # MatchPathPrefix calls.
+
+ # Heuristic 1: handle common prefixes for partitions
+ for c in COMMON_PREFIXES:
+ if not pathregex.startswith(c):
+ continue
+ found = False
+ for p in COMMON_PREFIXES[c]:
+ if prefix.startswith(p):
+ found = True
+ prefix = prefix[len(p):]
+ pathregex = pathregex[len(c):]
+ break
+ if not found:
+ return False
+
+ # Heuristic 2: compare normal characters as long as possible
+ idx = 0
+ while idx < len(prefix):
+ if idx == len(pathregex):
+ return False
+ if pathregex[idx] in fc_sort.META_CHARS or pathregex[idx] == '\\':
+ break
+ if pathregex[idx] != prefix[idx]:
+ return False
+ idx += 1
+ if idx == len(prefix):
+ return True
+
+ # Fall back to regex compile loop.
for i in range(len(pathregex), 0, -1):
try:
pattern = re.compile('^' + pathregex[0:i] + "$")