passcode

信息收集

在ssh连接靶机之后,看文件:

$ ls -ll
total 16
-r--r----- 1 root passcode_pwn 48 Jun 26 2014 flag
-r-xr-sr-x 1 root passcode_pwn 7485 Jun 26 2014 passcode
-rw-r--r-- 1 root root 858 Jun 26 2014 passcode.c

查看C文件:

$ cat passcode.c 
#include <stdio.h>
#include <stdlib.h>

void login(){
int passcode1;
int passcode2;

printf("enter passcode1 : ");
scanf("%d", passcode1); // lose '&'
fflush(stdin);

// ha! mommy told me that 32bit is vulnerable to bruteforcing :)
printf("enter passcode2 : ");
scanf("%d", passcode2); // lose '&'

printf("checking...\n");
if(passcode1==338150 && passcode2==13371337){
printf("Login OK!\n");
system("/bin/cat flag");
}
else{
printf("Login Failed!\n");
exit(0);
}
}

void welcome(){
char name[100];
printf("enter you name : ");
scanf("%100s", name); // lose '&'
printf("Welcome %s!\n", name);
}

int main(){
printf("Toddler's Secure Login System 1.0 beta.\n");

welcome();
login();

// something after login...
printf("Now I can safely trust you that you have credential :)\n");
return 0;
}

由于没有加地址引用,所以按逻辑输入会报错;

passcode@pwnable:~$ ./passcode 
Toddler's Secure Login System 1.0 beta.
enter you name : klose
Welcome klose!
enter passcode1 : 123
Segmentation fault (core dumped)
passcode@pwnable:~$

通过objdump来获取汇编信息;

	c7 04 24 24 9f 04 08 	movl   $0x8049f24,(%esp)
804855f: ff d0 call *%eax
8048561: c9 leave
8048562: c3 ret
8048563: 90 nop

08048564 <login>:
8048564: 55 push %ebp
8048565: 89 e5 mov %esp,%ebp
8048567: 83 ec 28 sub $0x28,%esp
804856a: b8 70 87 04 08 mov $0x8048770,%eax
804856f: 89 04 24 mov %eax,(%esp)
8048572: e8 a9 fe ff ff call 8048420 <printf@plt>
8048577: b8 83 87 04 08 mov $0x8048783,%eax
804857c: 8b 55 f0 mov -0x10(%ebp),%edx
804857f: 89 54 24 04 mov %edx,0x4(%esp)
8048583: 89 04 24 mov %eax,(%esp)
8048586: e8 15 ff ff ff call 80484a0 <__isoc99_scanf@plt>
804858b: a1 2c a0 04 08 mov 0x804a02c,%eax
8048590: 89 04 24 mov %eax,(%esp)
8048593: e8 98 fe ff ff call 8048430 <fflush@plt>
8048598: b8 86 87 04 08 mov $0x8048786,%eax
804859d: 89 04 24 mov %eax,(%esp)
80485a0: e8 7b fe ff ff call 8048420 <printf@plt>
80485a5: b8 83 87 04 08 mov $0x8048783,%eax
80485aa: 8b 55 f4 mov -0xc(%ebp),%edx
80485ad: 89 54 24 04 mov %edx,0x4(%esp)
80485b1: 89 04 24 mov %eax,(%esp)
80485b4: e8 e7 fe ff ff call 80484a0 <__isoc99_scanf@plt>
80485b9: c7 04 24 99 87 04 08 movl $0x8048799,(%esp)
80485c0: e8 8b fe ff ff call 8048450 <puts@plt>
80485c5: 81 7d f0 e6 28 05 00 cmpl $0x528e6,-0x10(%ebp)
80485cc: 75 23 jne 80485f1 <login+0x8d>
80485ce: 81 7d f4 c9 07 cc 00 cmpl $0xcc07c9,-0xc(%ebp)
80485d5: 75 1a jne 80485f1 <login+0x8d>
80485d7: c7 04 24 a5 87 04 08 movl $0x80487a5,(%esp)
80485de: e8 6d fe ff ff call 8048450 <puts@plt>
80485e3: c7 04 24 af 87 04 08 movl $0x80487af,(%esp)
80485ea: e8 71 fe ff ff call 8048460 <system@plt>
80485ef: c9 leave
80485f0: c3 ret
80485f1: c7 04 24 bd 87 04 08 movl $0x80487bd,(%esp)
80485f8: e8 53 fe ff ff call 8048450 <puts@plt>
80485fd: c7 04 24 00 00 00 00 movl $0x0,(%esp)
8048604: e8 77 fe ff ff call 8048480 <exit@plt>

08048609 <welcome>:
8048609: 55 push %ebp
804860a: 89 e5 mov %esp,%ebp
804860c: 81 ec 88 00 00 00 sub $0x88,%esp
8048612: 65 a1 14 00 00 00 mov %gs:0x14,%eax
8048618: 89 45 f4 mov %eax,-0xc(%ebp)
804861b: 31 c0 xor %eax,%eax
804861d: b8 cb 87 04 08 mov $0x80487cb,%eax
8048622: 89 04 24 mov %eax,(%esp)
8048625: e8 f6 fd ff ff call 8048420 <printf@plt>
804862a: b8 dd 87 04 08 mov $0x80487dd,%eax
804862f: 8d 55 90 lea -0x70(%ebp),%edx
8048632: 89 54 24 04 mov %edx,0x4(%esp)
8048636: 89 04 24 mov %eax,(%esp)
8048639: e8 62 fe ff ff call 80484a0 <__isoc99_scanf@plt>
804863e: b8 e3 87 04 08 mov $0x80487e3,%eax
8048643: 8d 55 90 lea -0x70(%ebp),%edx
8048646: 89 54 24 04 mov %edx,0x4(%esp)
804864a: 89 04 24 mov %eax,(%esp)
804864d: e8 ce fd ff ff call 8048420 <printf@plt>
8048652: 8b 45 f4 mov -0xc(%ebp),%eax
8048655: 65 33 05 14 00 00 00 xor %gs:0x14,%eax
804865c: 74 05 je 8048663 <welcome+0x5a>
804865e: e8 dd fd ff ff call 8048440 <__stack_chk_fail@plt>
8048663: c9 leave
8048664: c3 ret

08048665 <main>:
8048665: 55 push %ebp
8048666: 89 e5 mov %esp,%ebp
8048668: 83 e4 f0 and $0xfffffff0,%esp
804866b: 83 ec 10 sub $0x10,%esp
804866e: c7 04 24 f0 87 04 08 movl $0x80487f0,(%esp)
8048675: e8 d6 fd ff ff call 8048450 <puts@plt>
804867a: e8 8a ff ff ff call 8048609 <welcome>
804867f: e8 e0 fe ff ff call 8048564 <login>
8048684: c7 04 24 18 88 04 08 movl $0x8048818,(%esp)
804868b: e8 c0 fd ff ff call 8048450 <puts@plt>
8048690: b8 00 00 00 00 mov $0x0,%eax
8048695: c9 leave
8048696: c3 ret

可以看到:

804862a:	b8 dd 87 04 08       	mov    $0x80487dd,%eax
804862f: 8d 55 90 lea -0x70(%ebp),%edx
8048632: 89 54 24 04 mov %edx,0x4(%esp)
8048636: 89 04 24 mov %eax,(%esp)
8048639: e8 62 fe ff ff call 80484a0 <__isoc99_scanf@plt>

为name开辟了很大的空间,偏移在ebp-0x70;而passcode1在偏移ebp-0x10,passcode2在偏移ebp-0xc

因此尝试利用溢出进行覆盖;

pwn

338150转换成十六进制为:0x528E6

13371337转换成十六进制为:0xCC07C9

可以试着把C文件编译为可执行文件,进行调试;

gcc passcode.c -g -m32 -o passcode