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author | Qiuhao Li <Qiuhao.Li@outlook.com> | 2021-01-11 14:11:48 +0800 |
---|---|---|
committer | Thomas Huth <thuth@redhat.com> | 2021-01-11 14:59:21 +0100 |
commit | e72203abec8f15bb187c239256e7c991cb21601f (patch) | |
tree | 33a0b68e17886987be38f9664deb886fa03ecb18 /scripts/oss-fuzz | |
parent | 7b339f287fb73a628f0862823b880a6145faa6ec (diff) | |
download | qemu-e72203abec8f15bb187c239256e7c991cb21601f.zip qemu-e72203abec8f15bb187c239256e7c991cb21601f.tar.gz qemu-e72203abec8f15bb187c239256e7c991cb21601f.tar.bz2 |
fuzz: split write operand using binary approach
Currently, we split the write commands' data from the middle. If it does not
work, try to move the pivot left by one byte and retry until there is no
space.
But, this method has two flaws:
1. It may fail to trim all unnecessary bytes on the right side.
For example, there is an IO write command:
write addr uuxxxxuu
u is the unnecessary byte for the crash. Unlike ram write commands, in most
case, a split IO write won't trigger the same crash, So if we split from the
middle, we will get:
write addr uu (will be removed in next round)
write addr xxxxuu
For xxxxuu, since split it from the middle and retry to the leftmost byte
won't get the same crash, we will be stopped from removing the last two
bytes.
2. The algorithm complexity is O(n) since we move the pivot byte by byte.
To solve the first issue, we can try a symmetrical position on the right if
we fail on the left. As for the second issue, instead moving by one byte, we
can approach the boundary exponentially, achieving O(log(n)).
Give an example:
xxxxuu len=6
+
|
+
xxx,xuu 6/2=3 fail
+
+--------------+-------------+
| |
+ +
xx,xxuu 6/2^2=1 fail xxxxu,u 6-1=5 success
+ +
+------------------+----+ |
| | +-------------+ u removed
+ +
xx,xxu 5/2=2 fail xxxx,u 6-2=4 success
+
|
+-----------+ u removed
In some rare cases, this algorithm will fail to trim all unnecessary bytes:
xxxxxxxxxuxxxxxx
xxxxxxxx-xuxxxxxx Fail
xxxx-xxxxxuxxxxxx Fail
xxxxxxxxxuxx-xxxx Fail
...
I think the trade-off is worth it.
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: <SYCPR01MB3502D26F1BEB680CBBC169E5FCAB0@SYCPR01MB3502.ausprd01.prod.outlook.com>
Signed-off-by: Thomas Huth <thuth@redhat.com>
Diffstat (limited to 'scripts/oss-fuzz')
-rwxr-xr-x | scripts/oss-fuzz/minimize_qtest_trace.py | 29 |
1 files changed, 20 insertions, 9 deletions
diff --git a/scripts/oss-fuzz/minimize_qtest_trace.py b/scripts/oss-fuzz/minimize_qtest_trace.py index cacabf2..af9767f 100755 --- a/scripts/oss-fuzz/minimize_qtest_trace.py +++ b/scripts/oss-fuzz/minimize_qtest_trace.py @@ -97,7 +97,7 @@ def minimize_trace(inpath, outpath): prior = newtrace[i:i+remove_step] for j in range(i, i+remove_step): newtrace[j] = "" - print("Removing {lines} ...".format(lines=prior)) + print("Removing {lines} ...\n".format(lines=prior)) if check_if_trace_crashes(newtrace, outpath): i += remove_step # Double the number of lines to remove for next round @@ -110,9 +110,11 @@ def minimize_trace(inpath, outpath): 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) + if (newtrace[i].startswith("write") and not newtrace[i].startswith("write ")): suffix = newtrace[i].split()[0][-1] @@ -133,11 +135,15 @@ def minimize_trace(inpath, outpath): 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 - # middle does not work, try to move the pivot "left" and retry, until - # there is no space left. The idea is to prune unneccessary bytes from - # long writes, while accommodating arbitrary MemoryRegion access sizes - # and alignments. + # it into two separate write commands. If splitting the data operand + # from length/2^n bytes to the left does not work, try to move the pivot + # to the right side, then add one to n, until length/2^n == 0. The idea + # is to prune unneccessary bytes from long writes, while accommodating + # arbitrary MemoryRegion access sizes and alignments. + + # This algorithm will fail under some rare situations. + # e.g., xxxxxxxxxuxxxxxx (u is the unnecessary byte) + if newtrace[i].startswith("write "): addr = int(newtrace[i].split()[1], 16) length = int(newtrace[i].split()[2], 16) @@ -146,6 +152,7 @@ def minimize_trace(inpath, outpath): leftlength = int(length/2) rightlength = length - leftlength newtrace.insert(i+1, "") + power = 1 while leftlength > 0: newtrace[i] = "write {addr} {size} 0x{data}\n".format( addr=hex(addr), @@ -157,9 +164,13 @@ def minimize_trace(inpath, outpath): data=data[leftlength*2:]) if check_if_trace_crashes(newtrace, outpath): break - else: - leftlength -= 1 - rightlength += 1 + # move the pivot to right side + if leftlength < rightlength: + rightlength, leftlength = leftlength, rightlength + continue + power += 1 + leftlength = int(length/pow(2, power)) + rightlength = length - leftlength if check_if_trace_crashes(newtrace, outpath): i -= 1 else: |