qemu/scripts/oss-fuzz/minimize_qtest_trace.py

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#!/usr/bin/env python3
# -*- coding: utf-8 -*-
"""
This takes a crashing qtest trace and tries to remove superflous operations
"""
import sys
import os
import subprocess
import time
import struct
QEMU_ARGS = None
QEMU_PATH = None
TIMEOUT = 5
CRASH_TOKEN = None
# Minimization levels
M1 = False # try removing IO commands iteratively
M2 = False # try setting bits in operand of write/out to zero
write_suffix_lookup = {"b": (1, "B"),
"w": (2, "H"),
"l": (4, "L"),
"q": (8, "Q")}
def usage():
sys.exit("""\
Usage:
QEMU_PATH="/path/to/qemu" QEMU_ARGS="args" {} [Options] input_trace output_trace
By default, will try to use the second-to-last line in the output to identify
whether the crash occred. Optionally, manually set a string that idenitifes the
crash by setting CRASH_TOKEN=
Options:
-M1: enable a loop around the remove minimizer, which may help decrease some
timing dependant instructions. Off by default.
-M2: try setting bits in operand of write/out to zero. Off by default.
""".format((sys.argv[0])))
fuzz: accelerate non-crash detection We spend much time waiting for the timeout program during the minimization process until it passes a time limit. This patch hacks the CLOSED (indicates the redirection file closed) notification in QTest's output if it doesn't crash. Test with quadrupled trace input at: https://bugs.launchpad.net/qemu/+bug/1890333/comments/1 Original version: real 1m37.246s user 0m13.069s sys 0m8.399s Refined version: real 0m45.904s user 0m16.874s sys 0m10.042s Note: Sometimes the mutated or the same trace may trigger a different crash summary (second-to-last line) but indicates the same bug. For example, Bug 1910826 [1], which will trigger a stack overflow, may output summaries like: SUMMARY: AddressSanitizer: stack-overflow /home/qiuhao/hack/qemu/build/../softmmu/physmem.c:488 in flatview_do_translate or SUMMARY: AddressSanitizer: stack-overflow (/home/qiuhao/hack/qemu/build/qemu-system-i386+0x27ca049) in __asan_memcpy Etc. If we use the whole summary line as the token, we may be prevented from further minimization. So in this patch, we only use the first three words which indicate the type of crash: SUMMARY: AddressSanitizer: stack-overflow [1] https://bugs.launchpad.net/qemu/+bug/1910826 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: <SYCPR01MB350251DC04003450348FAF68FCAB0@SYCPR01MB3502.ausprd01.prod.outlook.com> Signed-off-by: Thomas Huth <thuth@redhat.com>
2021-01-11 09:11:46 +03:00
deduplication_note = """\n\
Note: While trimming the input, sometimes the mutated trace triggers a different
type crash but indicates the same bug. Under this situation, our minimizer is
incapable of recognizing and stopped from removing it. In the future, we may
use a more sophisticated crash case deduplication method.
\n"""
def check_if_trace_crashes(trace, path):
with open(path, "w") as tracefile:
tracefile.write("".join(trace))
rc = subprocess.Popen("timeout -s 9 {timeout}s {qemu_path} {qemu_args} 2>&1\
< {trace_path}".format(timeout=TIMEOUT,
qemu_path=QEMU_PATH,
qemu_args=QEMU_ARGS,
trace_path=path),
shell=True,
stdin=subprocess.PIPE,
fuzz: accelerate non-crash detection We spend much time waiting for the timeout program during the minimization process until it passes a time limit. This patch hacks the CLOSED (indicates the redirection file closed) notification in QTest's output if it doesn't crash. Test with quadrupled trace input at: https://bugs.launchpad.net/qemu/+bug/1890333/comments/1 Original version: real 1m37.246s user 0m13.069s sys 0m8.399s Refined version: real 0m45.904s user 0m16.874s sys 0m10.042s Note: Sometimes the mutated or the same trace may trigger a different crash summary (second-to-last line) but indicates the same bug. For example, Bug 1910826 [1], which will trigger a stack overflow, may output summaries like: SUMMARY: AddressSanitizer: stack-overflow /home/qiuhao/hack/qemu/build/../softmmu/physmem.c:488 in flatview_do_translate or SUMMARY: AddressSanitizer: stack-overflow (/home/qiuhao/hack/qemu/build/qemu-system-i386+0x27ca049) in __asan_memcpy Etc. If we use the whole summary line as the token, we may be prevented from further minimization. So in this patch, we only use the first three words which indicate the type of crash: SUMMARY: AddressSanitizer: stack-overflow [1] https://bugs.launchpad.net/qemu/+bug/1910826 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: <SYCPR01MB350251DC04003450348FAF68FCAB0@SYCPR01MB3502.ausprd01.prod.outlook.com> Signed-off-by: Thomas Huth <thuth@redhat.com>
2021-01-11 09:11:46 +03:00
stdout=subprocess.PIPE,
encoding="utf-8")
global CRASH_TOKEN
if CRASH_TOKEN is None:
fuzz: accelerate non-crash detection We spend much time waiting for the timeout program during the minimization process until it passes a time limit. This patch hacks the CLOSED (indicates the redirection file closed) notification in QTest's output if it doesn't crash. Test with quadrupled trace input at: https://bugs.launchpad.net/qemu/+bug/1890333/comments/1 Original version: real 1m37.246s user 0m13.069s sys 0m8.399s Refined version: real 0m45.904s user 0m16.874s sys 0m10.042s Note: Sometimes the mutated or the same trace may trigger a different crash summary (second-to-last line) but indicates the same bug. For example, Bug 1910826 [1], which will trigger a stack overflow, may output summaries like: SUMMARY: AddressSanitizer: stack-overflow /home/qiuhao/hack/qemu/build/../softmmu/physmem.c:488 in flatview_do_translate or SUMMARY: AddressSanitizer: stack-overflow (/home/qiuhao/hack/qemu/build/qemu-system-i386+0x27ca049) in __asan_memcpy Etc. If we use the whole summary line as the token, we may be prevented from further minimization. So in this patch, we only use the first three words which indicate the type of crash: SUMMARY: AddressSanitizer: stack-overflow [1] https://bugs.launchpad.net/qemu/+bug/1910826 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: <SYCPR01MB350251DC04003450348FAF68FCAB0@SYCPR01MB3502.ausprd01.prod.outlook.com> Signed-off-by: Thomas Huth <thuth@redhat.com>
2021-01-11 09:11:46 +03:00
try:
outs, _ = rc.communicate(timeout=5)
CRASH_TOKEN = " ".join(outs.splitlines()[-2].split()[0:3])
except subprocess.TimeoutExpired:
print("subprocess.TimeoutExpired")
return False
print("Identifying Crashes by this string: {}".format(CRASH_TOKEN))
global deduplication_note
print(deduplication_note)
return True
for line in iter(rc.stdout.readline, ""):
if "CLOSED" in line:
return False
if CRASH_TOKEN in line:
return True
fuzz: accelerate non-crash detection We spend much time waiting for the timeout program during the minimization process until it passes a time limit. This patch hacks the CLOSED (indicates the redirection file closed) notification in QTest's output if it doesn't crash. Test with quadrupled trace input at: https://bugs.launchpad.net/qemu/+bug/1890333/comments/1 Original version: real 1m37.246s user 0m13.069s sys 0m8.399s Refined version: real 0m45.904s user 0m16.874s sys 0m10.042s Note: Sometimes the mutated or the same trace may trigger a different crash summary (second-to-last line) but indicates the same bug. For example, Bug 1910826 [1], which will trigger a stack overflow, may output summaries like: SUMMARY: AddressSanitizer: stack-overflow /home/qiuhao/hack/qemu/build/../softmmu/physmem.c:488 in flatview_do_translate or SUMMARY: AddressSanitizer: stack-overflow (/home/qiuhao/hack/qemu/build/qemu-system-i386+0x27ca049) in __asan_memcpy Etc. If we use the whole summary line as the token, we may be prevented from further minimization. So in this patch, we only use the first three words which indicate the type of crash: SUMMARY: AddressSanitizer: stack-overflow [1] https://bugs.launchpad.net/qemu/+bug/1910826 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: <SYCPR01MB350251DC04003450348FAF68FCAB0@SYCPR01MB3502.ausprd01.prod.outlook.com> Signed-off-by: Thomas Huth <thuth@redhat.com>
2021-01-11 09:11:46 +03:00
print("\nWarning:")
print(" There is no 'CLOSED'or CRASH_TOKEN in the stdout of subprocess.")
print(" Usually this indicates a different type of crash.\n")
return False
# If previous write commands write the same length of data at the same
# interval, we view it as a hint.
def split_write_hint(newtrace, i):
HINT_LEN = 3 # > 2
if i <=(HINT_LEN-1):
return None
#find previous continuous write traces
k = 0
l = i-1
writes = []
while (k != HINT_LEN and l >= 0):
if newtrace[l].startswith("write "):
writes.append(newtrace[l])
k += 1
l -= 1
elif newtrace[l] == "":
l -= 1
else:
return None
if k != HINT_LEN:
return None
length = int(writes[0].split()[2], 16)
for j in range(1, HINT_LEN):
if length != int(writes[j].split()[2], 16):
return None
step = int(writes[0].split()[1], 16) - int(writes[1].split()[1], 16)
for j in range(1, HINT_LEN-1):
if step != int(writes[j].split()[1], 16) - \
int(writes[j+1].split()[1], 16):
return None
return (int(writes[0].split()[1], 16)+step, length)
def remove_lines(newtrace, outpath):
remove_step = 1
i = 0
while i < len(newtrace):
# 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] = ""
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>
2021-01-11 09:11:48 +03:00
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
remove_step *= 2
continue
# 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
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>
2021-01-11 09:11:48 +03:00
# 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)
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>
2021-01-11 09:11:48 +03:00
if (newtrace[i].startswith("write") and not
newtrace[i].startswith("write ")):
suffix = newtrace[i].split()[0][-1]
assert(suffix in write_suffix_lookup)
addr = int(newtrace[i].split()[1], 16)
value = int(newtrace[i].split()[2], 16)
for endianness in ['<', '>']:
data = struct.pack("{end}{size}".format(end=endianness,
size=write_suffix_lookup[suffix][1]),
value)
newtrace[i] = "write {addr} {size} 0x{data}\n".format(
addr=hex(addr),
size=hex(write_suffix_lookup[suffix][0]),
data=data.hex())
if(check_if_trace_crashes(newtrace, outpath)):
break
else:
newtrace[i] = prior[0]
# 3.) If it is a qtest write command: write addr len data, try to split
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>
2021-01-11 09:11:48 +03:00
# 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)
data = newtrace[i].split()[3][2:]
if length > 1:
# Can we get a hint from previous writes?
hint = split_write_hint(newtrace, i)
if hint is not None:
hint_addr = hint[0]
hint_len = hint[1]
if hint_addr >= addr and hint_addr+hint_len <= addr+length:
newtrace[i] = "write {addr} {size} 0x{data}\n".format(
addr=hex(hint_addr),
size=hex(hint_len),
data=data[(hint_addr-addr)*2:\
(hint_addr-addr)*2+hint_len*2])
if check_if_trace_crashes(newtrace, outpath):
# next round
i += 1
continue
newtrace[i] = prior[0]
# Try splitting it using a binary approach
leftlength = int(length/2)
rightlength = length - leftlength
newtrace.insert(i+1, "")
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>
2021-01-11 09:11:48 +03:00
power = 1
while leftlength > 0:
newtrace[i] = "write {addr} {size} 0x{data}\n".format(
addr=hex(addr),
size=hex(leftlength),
data=data[:leftlength*2])
newtrace[i+1] = "write {addr} {size} 0x{data}\n".format(
addr=hex(addr+leftlength),
size=hex(rightlength),
data=data[leftlength*2:])
if check_if_trace_crashes(newtrace, outpath):
break
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>
2021-01-11 09:11:48 +03:00
# 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:
newtrace[i] = prior[0]
del newtrace[i+1]
i += 1
def clear_bits(newtrace, outpath):
# try setting bits in operands of out/write to zero
i = 0
while i < len(newtrace):
if (not newtrace[i].startswith("write ") and not
newtrace[i].startswith("out")):
i += 1
continue
# write ADDR SIZE DATA
# outx ADDR VALUE
print("\nzero setting bits: {}".format(newtrace[i]))
prefix = " ".join(newtrace[i].split()[:-1])
data = newtrace[i].split()[-1]
data_bin = bin(int(data, 16))
data_bin_list = list(data_bin)
for j in range(2, len(data_bin_list)):
prior = newtrace[i]
if (data_bin_list[j] == '1'):
data_bin_list[j] = '0'
data_try = hex(int("".join(data_bin_list), 2))
# It seems qtest only accepts padded hex-values.
if len(data_try) % 2 == 1:
data_try = data_try[:2] + "0" + data_try[2:-1]
newtrace[i] = "{prefix} {data_try}\n".format(
prefix=prefix,
data_try=data_try)
if not check_if_trace_crashes(newtrace, outpath):
data_bin_list[j] = '1'
newtrace[i] = prior
i += 1
def minimize_trace(inpath, outpath):
global TIMEOUT
with open(inpath) as f:
trace = f.readlines()
start = time.time()
if not check_if_trace_crashes(trace, outpath):
sys.exit("The input qtest trace didn't cause a crash...")
end = time.time()
print("Crashed in {} seconds".format(end-start))
TIMEOUT = (end-start)*5
print("Setting the timeout for {} seconds".format(TIMEOUT))
newtrace = trace[:]
global M1, M2
# remove lines
old_len = len(newtrace) + 1
while(old_len > len(newtrace)):
old_len = len(newtrace)
print("trace lenth = ", old_len)
remove_lines(newtrace, outpath)
if not M1 and not M2:
break
newtrace = list(filter(lambda s: s != "", newtrace))
assert(check_if_trace_crashes(newtrace, outpath))
# set bits to zero
if M2:
clear_bits(newtrace, outpath)
assert(check_if_trace_crashes(newtrace, outpath))
if __name__ == '__main__':
if len(sys.argv) < 3:
usage()
if "-M1" in sys.argv:
M1 = True
if "-M2" in sys.argv:
M2 = True
QEMU_PATH = os.getenv("QEMU_PATH")
QEMU_ARGS = os.getenv("QEMU_ARGS")
if QEMU_PATH is None or QEMU_ARGS is None:
usage()
# if "accel" not in QEMU_ARGS:
# QEMU_ARGS += " -accel qtest"
CRASH_TOKEN = os.getenv("CRASH_TOKEN")
QEMU_ARGS += " -qtest stdio -monitor none -serial none "
minimize_trace(sys.argv[-2], sys.argv[-1])