安全漏洞本质扯谈之决战汇编代码

0x00 前言


当前比较流行也比较有效率的挖掘的漏洞的方法是Fuzzing,当然这也需要花点时间去写Fuzzing程序。然而不是每个东西都有必要去写Fuzzing程序,也不是每个东西都可以去Fuzzing的,所以也还是要继续修炼我们的二进制安全流派内功—“肉眼挖洞神功”。

安全漏洞在源代码层面和汇编代码层面会有不同的表现,想要进阶升级的话,必须清晰的了解每种安全漏洞类型在源代码层面和汇编代码层面都是张什么样的。有些安全漏洞在源代码层面容易发现它,有些安全漏洞则在汇编代码层面比较容易发现它。不管是源代码层面还是汇编代码层面,普通人看代码的时候即使看到有安全漏洞的那些代码的时候,往往是“只见代码,不见漏洞”。修炼“肉眼挖洞神功”的目标是看到那些疑似有问题的代码时候能够警觉,能够识别,并快速判定是否可形成安全漏洞。

早些时候给我们翰海源的小伙伴们讲过《深入安全漏洞-Root Cause of Vulnerabilities》和《决战汇编代码》,希望小伙伴们能够掌握安全漏洞的本质原因以及练就“肉眼挖洞神功”。今天把里面的部分例子跟大伙一起分享下。

0x01 源代码层面


1. Link Attack


*unix 下的 link attack,有意识到吗?

[cce]#!c
if(access(“file”,W_OK)!=0){
exit(1);
}
fd = open(“file”,O_WRONLY);

write(fd,buffer,sizeof(buffer));
[/cce]

2. Integer Overflow Example in OpenSSH 3.1


发生在真实的 openssh 3.1 ,有漏洞吗?

[cce]#!c
u_int nresp;

nresp = packet_get_int();
if(nresp){
response = xmalloc(nresp * sizeof(char*));
for(i=0; i3. Signed Integer Vulnerability Example


注意整形符号,能被绕过吗?

[cce]#!c
int read_user_data(int sockfd)
{
int length, sockfd, n;
char buffer[1024];

length = get_user_length(sockfd);

if(length > 1024){
error(“illegal input, not enough room in buffer\n”);
return 1;
}

if(read(sockfd, buffer, length) < 0){ error("read: %m"); return 1; } return 0; } [/cce]

4. Truncation Vulnerability Example in NFS


整形截断问题?

[cce]#!c
void assume_privs(unsigned short uid) {
seteuid(uid);
setuid(uid);
}

int become_user(int uid)
{
if (uid == 0)
die(“root isnt allowed”);
assume_privs(uid);
}
[/cce]

5. 苹果SSL/TLS 重大安全漏洞的细节


(CVE-2014-1266)多个goto fail造成重大安全隐患。

[cce]#!c
static OSStatus
SSLVerifySignedServerKeyExchange(SSLContext *ctx, bool isRsa, SSLBuffer signedParams,
uint8_t *signature, UInt16 signatureLen)
{

if ((err = ReadyHash(&SSLHashSHA1, &hashCtx)) != 0)
goto fail;
if ((err = SSLHashSHA1.update(&hashCtx, &clientRandom)) != 0)
goto fail;
if ((err = SSLHashSHA1.update(&hashCtx, &serverRandom)) != 0)
goto fail;
if ((err = SSLHashSHA1.update(&hashCtx, &signedParams)) != 0)
goto fail;
goto fail; <---- *** DANGER *** if ((err = SSLHashSHA1.final(&hashCtx, &hashOut)) != 0) goto fail; err = sslRawVerify(ctx, ctx->peerPubKey,
dataToSign, /* plaintext */
dataToSignLen, /* plaintext length */
signature,
signatureLen);

fail:
SSLFreeBuffer(&signedHashes);
SSLFreeBuffer(&hashCtx);
return err;

}
[/cce]

6. MS-RPC DCOM Buffer Overflow
(冲击波)


冲击波蠕虫只因为一个缓冲区溢出。

[cce]#!c
HRESULT GetMachineName (WCHAR * pwszPath) {
The WCHAR wszMachineName [N + 1]);

LPWSTR pwszServerName = wszMachineName;
while (* pwszPath! = L ‘\\’)
* PwszServerName + + = * pwszPath + +;

}
[/cce]

7. 有漏洞吗?


[cce]#!c
unsigned short read_length(int sockfd)
{
unsigned short len;

if(full_read(sockfd, (void *)&len, 2) != 2)
die(“could not read length!\n”);
return ntohs(len);
}

int read_packet(int sockfd)
{
struct header hdr;
short length;
char *buffer;

length = read_length(sockfd);

if(length > 1024){
error(“read_packet: length too large: %d\n”, length);
return 1;
}

buffer = (char *)malloc(length+1);

if((n = read(sockfd, buffer, length) < 0){ error("read: %m"); free(buffer); return 1; } buffer[n] = '\0'; return 0; } [/cce]

8. 有漏洞吗?


提示: 5rOo5oSPc216ZW9m

[cce]#!c
char *read_username(int sockfd)
{
char *buffer, *style, userstring[1024];
int i;

buffer = (char *)malloc(1024);

if(!buffer){
error(“buffer allocation failed: %m”);
return NULL;
}

if(read(sockfd, userstring, sizeof(userstring)-1) <= 0){ free(buffer); error("read failure: %m"); return NULL; } userstring[sizeof(userstring)-1] = '\0'; style = strchr(userstring, ':'); if(style) *style++ = '\0'; sprintf(buffer, "username=%.32s", userstring); if(style) snprintf(buffer, sizeof(buffer)-strlen(buffer)-1, ", style=%s\n", style); return buffer; } [/cce]

9. 有漏洞吗?


提示:

enter image description here

[cce]#!c
/* special thing for ldap.
* The parts are separated by question marks.
* From RFC 2255:
* ldapurl = scheme “://” [hostport] [“/”
* [dn [“?” [attributes] [“?” [scope]
* [“?” [filter] [“?” extensions]]]]]]
*/

if (!strncasecmp(uri, “ldap”, 4))
{
char *token[5];
int c = 0;

token[0] = cp = ap_pstrdup(p, cp);
while (*cp && c < 5) { if (*cp == '?') { token[++c] = cp + 1; *cp = '\0'; } ++cp; } [/cce]

10. 有漏洞吗?(Antisniff v1.1 Vulnerability)


tips

[cce]#!c
char *indx;
int count;
char nameStr[MAX_LEN]; //256

memset(nameStr, ‘\0’, sizeof(nameStr));

indx = (char *)(pkt + rr_offset);
count = (char)*indx;

while (count){
if (strlen(nameStr) + count < ( MAX_LEN - 1) ){ (char *)indx++; strncat(nameStr, (char *)indx, count); indx += count; count = (char)*indx; strncat(nameStr, ".", sizeof(nameStr) strlen(nameStr)); } else { fprintf(stderr, "Alert! Someone is attempting " "to send LONG DNS packets\n"); count = 0; } } nameStr[strlen(nameStr)-1] = '\0'; [/cce]

11. 还有漏洞吗?(Antisniff v1.1.1 Vulnerability)


[cce]#!c
char *indx;
int count;
char nameStr[MAX_LEN]; //256

memset(nameStr, ‘\0’, sizeof(nameStr));

indx = (char *)(pkt + rr_offset);
count = (char)*indx;

while (count){
/* typecast the strlen so we aren’t dependent on
the call to be properly setting to unsigned. */

if ((unsigned int)strlen(nameStr) +
(unsigned int)count < ( MAX_LEN - 1) ){ (char *)indx++; strncat(nameStr, (char *)indx, count); indx += count; count = (char)*indx; strncat(nameStr, ".", sizeof(nameStr) strlen(nameStr)); } else { fprintf(stderr, "Alert! Someone is attempting " "to send LONG DNS packets\n"); count = 0; } } nameStr[strlen(nameStr)-1] = '\0'; [/cce]

12. 还有漏洞吗?(Antisniff v1.1.2 Vulnerability)


[cce]#!c
unsigned char *indx;
unsigned int count;
unsigned char nameStr[MAX_LEN]; //256

memset(nameStr, ‘\0’, sizeof(nameStr));

indx = (char *)(pkt + rr_offset);
count = (char)*indx;

while (count){
if (strlen(nameStr) + count < ( MAX_LEN - 1) ){ indx++; strncat(nameStr, indx, count); indx += count; count = *indx; strncat(nameStr, ".", sizeof(nameStr) strlen(nameStr)); } else { fprintf(stderr, "Alert! Someone is attempting " "to send LONG DNS packets\n"); count = 0; } } nameStr[strlen(nameStr)-1] = '\0'; [/cce]

0x02 汇编代码层面


接下要在汇编代码堆里磨练下,下面的汇编代码有安全问题吗?都是些什么问题?

1. Safe or vulnerability?


[cce]#!bash
text:0040106B sub_40106B proc near ; CODE XREF: _main+58p
.text:0040106B
.text:0040106B var_10004 = byte ptr -10004h
.text:0040106B var_4 = dword ptr -4
.text:0040106B arg_0 = word ptr 8
.text:0040106B
.text:0040106B push ebp
.text:0040106C mov ebp, esp
.text:0040106E mov eax, 10004h
.text:00401073 call __alloca_probe
.text:00401078 mov eax, dword_404020
.text:0040107D xor eax, ebp
.text:0040107F mov [ebp+var_4], eax
.text:00401082 movsx eax, [ebp+arg_0]
.text:00401086 movsx eax, [ebp+eax+var_10004]
.text:0040108E push eax
.text:0040108F push offset Format ; “t %x”
.text:00401094 call ds:printf
.text:0040109A pop ecx
.text:0040109B pop ecx
.text:0040109C mov ecx, [ebp+var_4]
.text:0040109F xor ecx, ebp
.text:004010A1 xor eax, eax
.text:004010A3 call sub_401BD2
.text:004010A8 leave
[/cce]

2. Safe or vulnerability?


[cce]#!bash
text:004010AA sub_4010AA proc near ; CODE XREF: _main+60p
.text:004010AA
.text:004010AA var_190 = dword ptr -190h
.text:004010AA arg_0 = dword ptr 8
.text:004010AA
.text:004010AA push ebp
.text:004010AB mov ebp, esp
.text:004010AD mov eax, [ebp+arg_0]
.text:004010B0 sub esp, 190h
.text:004010B6 cmp eax, 64h
.text:004010B9 jle short loc_4010BE
.text:004010BB push 64h
.text:004010BD pop eax
.text:004010BE
.text:004010BE loc_4010BE: ; CODE XREF: sub_4010AA+Fj
.text:004010BE push [ebp+eax*4+var_190]
.text:004010C5 push offset Format ; “t %x”
.text:004010CA call ds:printf
.text:004010D0 pop ecx
[/cce]

3. Safe or vulnerability?


[cce]#!bash
.text:00401000 sub_401000 proc near ; CODE XREF: _main+48p
.text:00401000
.text:00401000 var_190 = dword ptr -190h
.text:00401000 arg_0 = dword ptr 8
.text:00401000
.text:00401000 push ebp
.text:00401001 mov ebp, esp
.text:00401003 mov eax, [ebp+arg_0]
.text:00401006 sub esp, 190h
.text:0040100C cmp eax, 64h
.text:0040100F jbe short loc_401014
.text:00401011 push 64h
.text:00401013 pop eax
.text:00401014
.text:00401014 loc_401014: ; CODE XREF: sub_401000+Fj
.text:00401014 push [ebp+eax*4+var_190]
.text:0040101B push offset Format ; “t %x”
.text:00401020 call ds:printf
.text:00401026 pop ecx
.text:00401027 pop ecx
[/cce]

4. Safe or vulnerability?


[cce]#!bash
.text:004010D6 sub_4010D6 proc near ; CODE XREF: _main+68p
.text:004010D6
.text:004010D6 arg_0 = dword ptr 4
.text:004010D6
.text:004010D6 push esi
.text:004010D7 push 64h ; Size
.text:004010D9 call ds:malloc
.text:004010DF mov esi, eax
.text:004010E1 mov eax, [esp+8+arg_0]
.text:004010E5 cmp eax, 64h
.text:004010E8 pop ecx
.text:004010E9 jle short loc_4010EE
.text:004010EB push 64h
.text:004010ED pop eax
.text:004010EE
.text:004010EE loc_4010EE: ; CODE XREF: sub_4010D6+13j
.text:004010EE movsx eax, byte ptr [eax+esi]
.text:004010F2 push eax
.text:004010F3 push offset Format ; “t %x”
.text:004010F8 call ds:printf
.text:004010FE push esi ; Memory
.text:004010FF call ds:free
[/cce]

5. Safe or vulnerability?


[cce]#!bash
. text:0040110C sub_40110C proc near ; CODE XREF: _main+70p
.text:0040110C
.text:0040110C var_19A0 = byte ptr -19A0h
.text:0040110C var_4 = dword ptr -4
.text:0040110C arg_0 = word ptr 8
.text:0040110C
.text:0040110C push ebp
.text:0040110D mov ebp, esp
.text:0040110F mov eax, 19A0h
.text:00401114 call __alloca_probe
.text:00401119 mov eax, dword_404020
.text:0040111E xor eax, ebp
.text:00401120 mov [ebp+var_4], eax
.text:00401123 movsx eax, [ebp+arg_0]
.text:00401127 movsx eax, [ebp+eax+var_19A0]
.text:0040112F push eax
.text:00401130 push offset Format ; “t %x”
.text:00401135 call ds:printf
.text:0040113B pop ecx
.text:0040113C pop ecx
.text:0040113D mov ecx, [ebp+var_4]
.text:00401140 xor ecx, ebp
[/cce]

6. Safe or vulnerability?


[cce]#!bash
.text:0040102C sub_40102C proc near ; CODE XREF: _main+50p
.text:0040102C
.text:0040102C var_10004 = byte ptr -10004h
.text:0040102C var_4 = dword ptr -4
.text:0040102C arg_0 = word ptr 8
.text:0040102C
.text:0040102C push ebp
.text:0040102D mov ebp, esp
.text:0040102F mov eax, 10004h
.text:00401034 call __alloca_probe
.text:00401039 mov eax, dword_404020
.text:0040103E xor eax, ebp
.text:00401040 mov [ebp+var_4], eax
.text:00401043 movzx eax, [ebp+arg_0]
.text:00401047 movsx eax, [ebp+eax+var_10004]
.text:0040104F push eax
.text:00401050 push offset Format ; “t %x”
.text:00401055 call ds:printf
.text:0040105B pop ecx
[/cce]

7. Safe or vulnerability?


[cce]#!bash
. .text:0040118A ; int __cdecl sub_40118A(LPCSTR lpMultiByteStr)
.text:0040118A sub_40118A proc near ; CODE XREF: _main+91p
.text:0040118A
.text:0040118A WideCharStr = word ptr -44h
.text:0040118A var_4 = dword ptr -4
.text:0040118A lpMultiByteStr = dword ptr 8
.text:0040118A
.text:0040118A push ebp
.text:0040118B mov ebp, esp
.text:0040118D sub esp, 44h
.text:00401190 mov eax, dword_404020
.text:00401195 xor eax, ebp
.text:00401197 mov [ebp+var_4], eax
.text:0040119A mov ecx, [ebp+lpMultiByteStr]
.text:0040119D mov eax, ecx
.text:0040119F push esi
.text:004011A0 lea esi, [eax+1]
.text:004011A3
.text:004011A3 loc_4011A3: ; CODE XREF: sub_40118A+1Ej
.text:004011A3 mov dl, [eax]
.text:004011A5 inc eax
.text:004011A6 test dl, dl
.text:004011A8 jnz short loc_4011A3
.text:004011AA push 40h ; cchWideChar
.text:004011AC lea edx, [ebp+WideCharStr]
.text:004011AF push edx ; lpWideCharStr
.text:004011B0 sub eax, esi
.text:004011B2 push eax ; cbMultiByte
.text:004011B3 push ecx ; lpMultiByteStr
.text:004011B4 push 0 ; dwFlags
.text:004011B6 push 0 ; CodePage
.text:004011B8 call ds:MultiByteToWideChar
[/cce]

8. Safe or vulnerability?


[cce]#!bash
text:00401738 Size = dword ptr 4
.text:00401738 Src = dword ptr 8
.text:00401738
.text:00401738 push esi
.text:00401739 push edi
.text:0040173A mov edi, [esp+8+Size]
.text:0040173E lea eax, [edi+1]
.text:00401741 push eax ; Size
.text:00401742 call ds:__imp__malloc
.text:00401748 mov esi, eax
.text:0040174A test esi, esi
.text:0040174C pop ecx
.text:0040174D jz short loc_401774
.text:0040174F push edi ; Size
.text:00401750 push [esp+0Ch+Src] ; Src
.text:00401754 push esi ; Dst
.text:00401755 call _memcpy
.text:0040175A push esi
.text:0040175B push offset aS ; “%s\n”
.text:00401760 mov byte ptr [esi+edi], 0
.text:00401764 call ds:__imp__printf
[/cce]

0x04 最后


汇编代码层面的例子需要想想它们的源代码都是怎么写的才能编译出来这些有安全漏洞的代码。做到源代码层面和汇编代码层面通吃。更要做到从宏观上了解程序世界功能和规则,从微观上了解其实现细节。

上面的例子都不到一百行的汇编代码,都找出问题了吗?决战汇编代码就是需要我们在真实程序世界的上千万行汇编代码中找出那些隐藏的安全漏洞,这个难度会高很多。其中有对抗有编程者的防御措施,需要思考如何打败它,是和程序世界的缔造者们的一次跨时间和空间无声的较量。

以上这些其实是一个引子,需要小伙伴继续深入研究推演和锻炼。推荐学习Mark Down & JohnMcDonald两位大牛的书《The Art of Software Security Assessment– Identifying andavoiding software vulnerabilities》(某些例子来自该书),希望对小伙伴们有所帮助。

Comment

评论:虽然看不明白,但是感觉很厉害的样子

评论:虽然看不明白,但是感觉很厉害的样子

评论:求详解。

评论:虽然看不明白,但是感觉很厉害的样子

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