There is a Lua script that is debugged, most likely with XFuscator, is it possible to deobuscate and get the source code?

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Part of the code

local ________ = ____[#____ - 9]("loadstring(\"return ____\")()") local __ = { _(54), _(152), _(234), _(194), _(162), _(0), _(2), _(8)} local ________ = ____[3] return ____[11](assert(#____[8](__) == 15889 and ____[8](__) or nil, "Tampering detected"), ____[#____])()
  • You can edit the question text by clicking edit below the question text. - aleksandr barakin
  • @EgorSkriptunoff, in stackoverflow.com/questions/15961156/lua-code-deobfuscation wrote about some possibility of deobfuscation. You can describe in more detail how to write a file in a bitcode, and then decompile it into readable Lua code (for example, an obfuscated 123.lua file)? Maybe something so doing. - Maxim
  • XFuscator builds code that makes meaningless distracting manipulations of variables, and then simply loads bytecode. You don’t have to poke around in this distracting code - just intercept the call to the loadstring and get the bytecode in finished form. The difficulty in understanding bytecode. Call luac -l -l файл_с_байт-кодом and stykayte. Bytecode is not compatible with other versions of Lua, and this is a bad XFuscator. You need to hook into a bytecode on the same version (the Lua version is determined by the fifth byte of the bytecode: 0x51 = Lua 5.1). For Lua 5.1, there is a decompiler that builds a more readable Lua source by bytecode. - Egor Skriptunoff
  • Lua 5.1 Decompiler - Egor Skriptunoff
  • @Maxim - via @Username you can send alerts only to those who posted something in this thread. This is how the SO engine works. Alas, it is not possible to call someone from the side. - Egor Skriptunoff

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