2020-10-23 18:07:40 +03:00
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#!/usr/bin/env python3
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# -*- coding: utf-8 -*-
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"""
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This takes a crashing qtest trace and tries to remove superflous operations
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"""
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import sys
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import os
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import subprocess
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import time
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import struct
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QEMU_ARGS = None
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QEMU_PATH = None
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TIMEOUT = 5
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CRASH_TOKEN = None
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2021-01-11 09:11:51 +03:00
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# Minimization levels
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M1 = False # try removing IO commands iteratively
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M2 = False # try setting bits in operand of write/out to zero
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2020-10-23 18:07:40 +03:00
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write_suffix_lookup = {"b": (1, "B"),
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"w": (2, "H"),
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"l": (4, "L"),
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"q": (8, "Q")}
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def usage():
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sys.exit("""\
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2021-01-11 09:11:51 +03:00
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Usage:
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QEMU_PATH="/path/to/qemu" QEMU_ARGS="args" {} [Options] input_trace output_trace
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2020-10-23 18:07:40 +03:00
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By default, will try to use the second-to-last line in the output to identify
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whether the crash occred. Optionally, manually set a string that idenitifes the
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crash by setting CRASH_TOKEN=
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2021-01-11 09:11:51 +03:00
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Options:
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-M1: enable a loop around the remove minimizer, which may help decrease some
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timing dependant instructions. Off by default.
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-M2: try setting bits in operand of write/out to zero. Off by default.
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2020-10-23 18:07:40 +03:00
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""".format((sys.argv[0])))
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2021-01-11 09:11:46 +03:00
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deduplication_note = """\n\
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Note: While trimming the input, sometimes the mutated trace triggers a different
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type crash but indicates the same bug. Under this situation, our minimizer is
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incapable of recognizing and stopped from removing it. In the future, we may
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use a more sophisticated crash case deduplication method.
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\n"""
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2020-10-23 18:07:40 +03:00
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def check_if_trace_crashes(trace, path):
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with open(path, "w") as tracefile:
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tracefile.write("".join(trace))
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rc = subprocess.Popen("timeout -s 9 {timeout}s {qemu_path} {qemu_args} 2>&1\
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< {trace_path}".format(timeout=TIMEOUT,
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qemu_path=QEMU_PATH,
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qemu_args=QEMU_ARGS,
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trace_path=path),
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shell=True,
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stdin=subprocess.PIPE,
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2021-01-11 09:11:46 +03:00
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stdout=subprocess.PIPE,
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encoding="utf-8")
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global CRASH_TOKEN
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2020-10-23 18:07:40 +03:00
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if CRASH_TOKEN is None:
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2021-01-11 09:11:46 +03:00
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try:
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outs, _ = rc.communicate(timeout=5)
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CRASH_TOKEN = " ".join(outs.splitlines()[-2].split()[0:3])
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except subprocess.TimeoutExpired:
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print("subprocess.TimeoutExpired")
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return False
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print("Identifying Crashes by this string: {}".format(CRASH_TOKEN))
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global deduplication_note
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print(deduplication_note)
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return True
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for line in iter(rc.stdout.readline, ""):
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if "CLOSED" in line:
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return False
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if CRASH_TOKEN in line:
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return True
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2020-10-23 18:07:40 +03:00
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2021-01-11 09:11:46 +03:00
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print("\nWarning:")
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print(" There is no 'CLOSED'or CRASH_TOKEN in the stdout of subprocess.")
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print(" Usually this indicates a different type of crash.\n")
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return False
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2020-10-23 18:07:40 +03:00
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2021-01-11 09:11:52 +03:00
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# If previous write commands write the same length of data at the same
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# interval, we view it as a hint.
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def split_write_hint(newtrace, i):
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HINT_LEN = 3 # > 2
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if i <=(HINT_LEN-1):
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return None
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#find previous continuous write traces
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k = 0
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l = i-1
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writes = []
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while (k != HINT_LEN and l >= 0):
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if newtrace[l].startswith("write "):
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writes.append(newtrace[l])
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k += 1
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l -= 1
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elif newtrace[l] == "":
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l -= 1
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else:
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return None
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if k != HINT_LEN:
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return None
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length = int(writes[0].split()[2], 16)
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for j in range(1, HINT_LEN):
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if length != int(writes[j].split()[2], 16):
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return None
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step = int(writes[0].split()[1], 16) - int(writes[1].split()[1], 16)
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for j in range(1, HINT_LEN-1):
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if step != int(writes[j].split()[1], 16) - \
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int(writes[j+1].split()[1], 16):
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return None
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return (int(writes[0].split()[1], 16)+step, length)
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2021-01-11 09:11:49 +03:00
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def remove_lines(newtrace, outpath):
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2021-01-11 09:11:47 +03:00
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remove_step = 1
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2021-01-11 09:11:49 +03:00
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i = 0
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2020-10-23 18:07:40 +03:00
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while i < len(newtrace):
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2021-01-11 09:11:47 +03:00
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# 1.) Try to remove lines completely and reproduce the crash.
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# If it works, we're done.
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if (i+remove_step) >= len(newtrace):
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remove_step = 1
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prior = newtrace[i:i+remove_step]
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for j in range(i, i+remove_step):
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newtrace[j] = ""
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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
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xxx,xuu 6/2=3 fail
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+--------------+-------------+
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xx,xxuu 6/2^2=1 fail xxxxu,u 6-1=5 success
+ +
+------------------+----+ |
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+ +
xx,xxu 5/2=2 fail xxxx,u 6-2=4 success
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+-----------+ 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
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print("Removing {lines} ...\n".format(lines=prior))
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2020-10-23 18:07:40 +03:00
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if check_if_trace_crashes(newtrace, outpath):
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2021-01-11 09:11:47 +03:00
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i += remove_step
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# Double the number of lines to remove for next round
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remove_step *= 2
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2020-10-23 18:07:40 +03:00
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continue
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2021-01-11 09:11:47 +03:00
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# Failed to remove multiple IOs, fast recovery
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if remove_step > 1:
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for j in range(i, i+remove_step):
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newtrace[j] = prior[j-i]
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remove_step = 1
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continue
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newtrace[i] = prior[0] # remove_step = 1
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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
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+
xxx,xuu 6/2=3 fail
+
+--------------+-------------+
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+ +
xx,xxuu 6/2^2=1 fail xxxxu,u 6-1=5 success
+ +
+------------------+----+ |
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+ +
xx,xxu 5/2=2 fail xxxx,u 6-2=4 success
+
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+-----------+ 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
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2020-10-23 18:07:40 +03:00
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# 2.) Try to replace write{bwlq} commands with a write addr, len
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# command. Since this can require swapping endianness, try both LE and
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# BE options. We do this, so we can "trim" the writes in (3)
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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
+
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+
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
+
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+-----------+ 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
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2020-10-23 18:07:40 +03:00
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if (newtrace[i].startswith("write") and not
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newtrace[i].startswith("write ")):
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suffix = newtrace[i].split()[0][-1]
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assert(suffix in write_suffix_lookup)
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addr = int(newtrace[i].split()[1], 16)
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value = int(newtrace[i].split()[2], 16)
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for endianness in ['<', '>']:
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data = struct.pack("{end}{size}".format(end=endianness,
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size=write_suffix_lookup[suffix][1]),
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value)
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newtrace[i] = "write {addr} {size} 0x{data}\n".format(
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addr=hex(addr),
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size=hex(write_suffix_lookup[suffix][0]),
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data=data.hex())
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if(check_if_trace_crashes(newtrace, outpath)):
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break
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else:
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2021-01-11 09:11:47 +03:00
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newtrace[i] = prior[0]
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2020-10-23 18:07:40 +03:00
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# 3.) If it is a qtest write command: write addr len data, try to split
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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
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# it into two separate write commands. If splitting the data operand
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# from length/2^n bytes to the left does not work, try to move the pivot
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# to the right side, then add one to n, until length/2^n == 0. The idea
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# is to prune unneccessary bytes from long writes, while accommodating
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# arbitrary MemoryRegion access sizes and alignments.
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# This algorithm will fail under some rare situations.
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# e.g., xxxxxxxxxuxxxxxx (u is the unnecessary byte)
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2020-10-23 18:07:40 +03:00
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if newtrace[i].startswith("write "):
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addr = int(newtrace[i].split()[1], 16)
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length = int(newtrace[i].split()[2], 16)
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data = newtrace[i].split()[3][2:]
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if length > 1:
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2021-01-11 09:11:52 +03:00
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# Can we get a hint from previous writes?
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hint = split_write_hint(newtrace, i)
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if hint is not None:
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hint_addr = hint[0]
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hint_len = hint[1]
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if hint_addr >= addr and hint_addr+hint_len <= addr+length:
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newtrace[i] = "write {addr} {size} 0x{data}\n".format(
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addr=hex(hint_addr),
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size=hex(hint_len),
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data=data[(hint_addr-addr)*2:\
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(hint_addr-addr)*2+hint_len*2])
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if check_if_trace_crashes(newtrace, outpath):
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# next round
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i += 1
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continue
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newtrace[i] = prior[0]
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# Try splitting it using a binary approach
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2020-10-23 18:07:40 +03:00
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leftlength = int(length/2)
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rightlength = length - leftlength
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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
|
2020-10-23 18:07:40 +03:00
|
|
|
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
|
2020-10-23 18:07:40 +03:00
|
|
|
if check_if_trace_crashes(newtrace, outpath):
|
|
|
|
i -= 1
|
|
|
|
else:
|
2021-01-11 09:11:47 +03:00
|
|
|
newtrace[i] = prior[0]
|
2020-10-23 18:07:40 +03:00
|
|
|
del newtrace[i+1]
|
|
|
|
i += 1
|
2021-01-11 09:11:49 +03:00
|
|
|
|
|
|
|
|
2021-01-11 09:11:50 +03:00
|
|
|
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
|
|
|
|
|
|
|
|
|
2021-01-11 09:11:49 +03:00
|
|
|
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[:]
|
2021-01-11 09:11:51 +03:00
|
|
|
global M1, M2
|
2021-01-11 09:11:49 +03:00
|
|
|
|
|
|
|
# remove lines
|
|
|
|
old_len = len(newtrace) + 1
|
|
|
|
while(old_len > len(newtrace)):
|
|
|
|
old_len = len(newtrace)
|
2021-01-11 09:11:51 +03:00
|
|
|
print("trace lenth = ", old_len)
|
2021-01-11 09:11:49 +03:00
|
|
|
remove_lines(newtrace, outpath)
|
2021-01-11 09:11:51 +03:00
|
|
|
if not M1 and not M2:
|
|
|
|
break
|
2021-01-11 09:11:49 +03:00
|
|
|
newtrace = list(filter(lambda s: s != "", newtrace))
|
2021-01-11 09:11:50 +03:00
|
|
|
assert(check_if_trace_crashes(newtrace, outpath))
|
2021-01-11 09:11:49 +03:00
|
|
|
|
2021-01-11 09:11:50 +03:00
|
|
|
# set bits to zero
|
2021-01-11 09:11:51 +03:00
|
|
|
if M2:
|
|
|
|
clear_bits(newtrace, outpath)
|
2021-01-11 09:11:49 +03:00
|
|
|
assert(check_if_trace_crashes(newtrace, outpath))
|
2020-10-23 18:07:40 +03:00
|
|
|
|
|
|
|
|
|
|
|
if __name__ == '__main__':
|
|
|
|
if len(sys.argv) < 3:
|
|
|
|
usage()
|
2021-01-11 09:11:51 +03:00
|
|
|
if "-M1" in sys.argv:
|
|
|
|
M1 = True
|
|
|
|
if "-M2" in sys.argv:
|
|
|
|
M2 = True
|
2020-10-23 18:07:40 +03:00
|
|
|
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 "
|
2021-01-11 09:11:51 +03:00
|
|
|
minimize_trace(sys.argv[-2], sys.argv[-1])
|