taultunleashed

It's actually really quite simple.
Code:

if ( !IsDebuggerPresent() ) { v51 = 15; v52 = 0; LOBYTE(v53) = 0; sub_421910((int)&v54, "HttpPatchFolder", 15u); v21 = sub_5427E0(); v56 = *(_DWORD *)(sub_543D80(&v55, &v54, v21) + 20) != 0; if ( v57 >= 16 ) sub_42A1F0(); v57 = 15; v58 = 0; LOBYTE(v59) = 0; if ( v51 >= 16 ) sub_42A1F0(); if ( v56 ) { v56 = sub_5B1670(); v11 = sub_5B1A90(); v60 = v11; if ( (!v56 || !v11) && (sub_F1AEA0("bValidClientHash && bValidPatcherKey", ".\Main.cpp", 238, 1), !v60) || !v56 ) { ShellExecuteA(0, "open", "ConanPatcher.exe", &Parameters, 0, 1); ExitProcess(0); } } }

Which basically says "Skip hash check if we are being debugged." The more complete approach however is this: 1. Get current 64-bit system time: __time64_t cur_time = _time64(0); 2. use TEA (hxxp://en.wikipedia.org/wiki/Tiny_Encryption_Algorithm) to encrypt with your key = {0x5BCE568E, 0x0FB2E8CBC, 0x0A324E6D8, 0x0F22BF865} and delta = 0x61C886470 3. Convert value from step 2 to a hex string. 4. Use process explorer to get the value of -clienthash 5. Start the process: exec AgeOfConan.exe -clienthash -key For anyone who found this as annoying as I did, you're welcome.

Can members say yay or nay to this so we may give tu bucks and/or premium. If you nay the submission remember to include as to why.

well then first i want to know why i should let aoc start without a patcher :S.. as i think, correct if im wrong, if u dont patch aoc.. u cant play it:) whats the use of this no patcher run? i mean this in a positiv way, i just wanna know .. ty *noob and stuff*

oww.. ye i get that sometimes.. it dloads a random 32 mb patch or sumthing.. but i dont know how to write/ or where to write is thus i cannot try it... but looks pretty good to me

some people launch aoc more than once every time they patch. it's useful for skipping that little bit of time you have to waste on the patcher every time. if you can't login due to client version, you know to run the patcher.

as for the actual original post, i'm not sure what type of code that is, so i can not confirm nor deny.