Bladeren bron

fuzz: double the IOs to remove for every loop

Instead of removing IO instructions one by one, we can try deleting multiple
instructions at once. According to the locality of reference, we double the
number of instructions to remove for the next round and recover it to one
once we fail.

This patch is usually significant for large input.

Test with quadrupled trace input at:
  https://bugs.launchpad.net/qemu/+bug/1890333/comments/1

Patched 1/6 version:
  real  0m45.904s
  user  0m16.874s
  sys   0m10.042s

Refined version:
  real  0m11.412s
  user  0m6.888s
  sys   0m3.325s

Signed-off-by: Qiuhao Li <Qiuhao.Li@outlook.com>
Reviewed-by: Alexander Bulekov <alxndr@bu.edu>
Tested-by: Alexander Bulekov <alxndr@bu.edu>
Message-Id: <SYCPR01MB350280A67BB55C3FADF173E3FCAB0@SYCPR01MB3502.ausprd01.prod.outlook.com>
Signed-off-by: Thomas Huth <thuth@redhat.com>
Qiuhao Li 4 jaren geleden
bovenliggende
commit
7b339f287f
1 gewijzigde bestanden met toevoegingen van 21 en 12 verwijderingen
  1. 21 12
      scripts/oss-fuzz/minimize_qtest_trace.py

+ 21 - 12
scripts/oss-fuzz/minimize_qtest_trace.py

@@ -88,19 +88,28 @@ def minimize_trace(inpath, outpath):
 
     i = 0
     newtrace = trace[:]
-    # For each line
+    remove_step = 1
     while i < len(newtrace):
-        # 1.) Try to remove it completely and reproduce the crash. If it works,
-        # we're done.
-        prior = newtrace[i]
-        print("Trying to remove {}".format(newtrace[i]))
-        # Try to remove the line completely
-        newtrace[i] = ""
+        # 1.) Try to remove lines completely and reproduce the crash.
+        # If it works, we're done.
+        if (i+remove_step) >= len(newtrace):
+            remove_step = 1
+        prior = newtrace[i:i+remove_step]
+        for j in range(i, i+remove_step):
+            newtrace[j] = ""
+        print("Removing {lines} ...".format(lines=prior))
         if check_if_trace_crashes(newtrace, outpath):
-            i += 1
+            i += remove_step
+            # Double the number of lines to remove for next round
+            remove_step *= 2
             continue
-        newtrace[i] = prior
-
+        # Failed to remove multiple IOs, fast recovery
+        if remove_step > 1:
+            for j in range(i, i+remove_step):
+                newtrace[j] = prior[j-i]
+            remove_step = 1
+            continue
+        newtrace[i] = prior[0] # remove_step = 1
         # 2.) Try to replace write{bwlq} commands with a write addr, len
         # command. Since this can require swapping endianness, try both LE and
         # BE options. We do this, so we can "trim" the writes in (3)
@@ -121,7 +130,7 @@ def minimize_trace(inpath, outpath):
                 if(check_if_trace_crashes(newtrace, outpath)):
                     break
             else:
-                newtrace[i] = prior
+                newtrace[i] = prior[0]
 
         # 3.) If it is a qtest write command: write addr len data, try to split
         # it into two separate write commands. If splitting the write down the
@@ -154,7 +163,7 @@ def minimize_trace(inpath, outpath):
                 if check_if_trace_crashes(newtrace, outpath):
                     i -= 1
                 else:
-                    newtrace[i] = prior
+                    newtrace[i] = prior[0]
                     del newtrace[i+1]
         i += 1
     check_if_trace_crashes(newtrace, outpath)