How does this milw0rm heap spraying exploit work?

Question

I usually do not have difficulty to read JavaScript code but for this one I can’t figure out the logic. The code is from an exploit that has been published 4 days ago. You c

Answer 1

Simple shellcode example

Hello world in assembly at&t syntax x86 I believe (Wizard in Training).

set up the file:vim shellcodeExample.s

.text           #required
.goblal _start  #required

_start:         #main function
 jmp one        #jump to the section labeled one:

two:
 pop  %rcx         #pop %rcx off the stack, or something
 xor  %rax, %rax   #Clear
 movl 4, %rax      #use sys_write(printf || std::cout)
 xor  %rbx, %rbx   #Clear
 inc  %rbx         #increment %rbx to 1 stdout(terminal)
 xor  %rdx, %rdx   #Clear Registers or something
 movb $13, %dl     #String Size
 int  $0x80

one:
 call two                   #jump up to section two:
 .ascii "Hello World\r\n"   #make the string one of the starting memory 
                            #^-addresses

compile like so:as -o shellcodeExample.o shellcodeExample.s ; ld -s -o shellcode shellcodeExample.o

Now you have a binary that prints out hello world. to convert the binary into shell code type in: objdump -D shellcode

you will get the output:

shellcode:     file format elf64-x86-64


Disassembly of section .text:

0000000000400078 <.text>:
  400078:   eb 1a                   jmp    0x400094
  40007a:   59                      pop    %rcx
  40007b:   48 31 c0                xor    %rax,%rax
  40007e:   b0 04                   mov    $0x4,%al
  400080:   48 31 db                xor    %rbx,%rbx
  400083:   48 ff c3                inc    %rbx
  400086:   48 31 d2                xor    %rdx,%rdx
  400089:   b2 0d                   mov    $0xd,%dl
  40008b:   cd 80                   int    $0x80
  40008d:   b0 01                   mov    $0x1,%al
  40008f:   48 ff cb                dec    %rbx
  400092:   cd 80                   int    $0x80
  400094:   e8 e1 ff ff ff          callq  0x40007a
  400099:   68 65 6c 6c 6f          pushq  $0x6f6c6c65
  40009e:   20 77 6f                and    %dh,0x6f(%rdi)
  4000a1:   72 6c                   jb     0x40010f
  4000a3:   64                      fs
  4000a4:   0d                      .byte 0xd
  4000a5:   0a                      .byte 0xa

Now if you look on the 4th line with text you will see: 400078: eb 1a jmp 0x400094

the part that says eb 1a is the hexadecimal representation of the assembly instruction jmp one where "one" is the memory address of your string.

to prep your shellcode for execution open up another text file and store the hex values in a character array. To format the shell code correctly you type in a \x before every hex value.

the upcoming shell code example will look like the following according to the objdump command output:

unsigned char PAYLOAD[] = 
"\xeb\x1a\x59\x48\x31\xc0\xb0\x04\x48\x31\xdb\x48\xff\xc3\x48\x31\xd2\xb2\xd0\xcd\x80\xb0\x01\x48\xff\xcb\xcd\x80\xe8\xe1\xff\xff\xff\x68\x65\x6c\x6c\x6f\x20\x77\x6f\x72\x6c\x64\x0d\x0a";

This example uses C for the array. Now you have working shellcode that will write to stdout "hello world"

you can test the shell code by placing it in a vulnerability or you can write the following c program to test it:

vim execShellcode.cc; //linux command to create c file.

/*Below is the content of execShellcode.cc*/
unsigned char PAYLOAD[] = 
"\xeb\x1a\x59\x48\x31\xc0\xb0\x04\x48\x31\xdb\x48\xff\xc3\x48\x31\xd2\xb2\xd0\xcd\x80\xb0\x01\x48\xff\xcb\xcd\x80\xe8\xe1\xff\xff\xff\x68\x65\x6c\x6c\x6f\x20\x77\x6f\x72\x6c\x64\x0d\x0a";

int main(){
    ((void(*)(void))PAYLOAD)();
    return 0;
}

To compile the program type in:

gcc -fno-stack-protector -z execstack execShellcode.cc -o run

run with ./run You know have a working example of simple shellcode development that was tested in linux mint/debian.