Decrypt a code (so I understand it is in ASCII)

Question

Decrypt a code (so I understand it is in ASCII)

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#1 by (1 votes)

1

Well, basically I would like to know if there is any way to transform a script written in ASCII into Characters and if so, how would I transform it?

Code (1)

 local code = '7  minimize = true -- minimizar a mainbp?
  bps = 1 -- quantidade de bps, deixar em 8
  local lastTime = os.clock()
  local main = Container.GetFirst()
  Module.New("reconnect", function()   
  if (os.difftime(os.clock(), lastTime) >= 1) then      
  local ping = Self.Ping()        
  while (#Container.GetAll() < bps) do      
  Cavebot.Stop()        
  Self.CloseContainers()        
  wait(2000)       
  Self.OpenMainBackpack(minimize)            
  wait(ping + 1000)     
  Cavebot.Start()           
  end    
  end    
  lastTime = os.clock()
  end)
  local script = string.dump(
  function()
  minimize = true -- minimizar a mainbp?
  bps = 1 -- quantidade de bps, deixar em 8
  local lastTime = os.clock()
  local main = Container.GetFirst()
  Module.New("reconnect", function()
  if (os.difftime(os.clock(), lastTime) >= 1) then
  local ping = Self.Ping()
  while (#Container.GetAll() < bps) do  
  Cavebot.Stop()
  Self.CloseContainers()    
  wait(2000)
  Self.OpenMainBackpack(minimize)    
  wait(ping + 1000)     
  Cavebot.Start()     
  end
  end
  lastTime = os.clock()
  end)

  end
  )

  buff=""
  for v=1,string.len(script) do --Convert our string into a hex string.
  buff=buff..'\'..string.byte(script,v)
  end

  file=io.open('encrypted.txt','w') --Output our bytecode into ascii format to encrypted.txt
  file:write(buff)
  file:flush()
  file:close()
 local code = '7  minimize = true -- minimizar a mainbp?
  bps = 1 -- quantidade de bps, deixar em 8
  local lastTime = os.clock()
  local main = Container.GetFirst()
  Module.New("reconnect", function()   
  if (os.difftime(os.clock(), lastTime) >= 1) then      
  local ping = Self.Ping()        
  while (#Container.GetAll() < bps) do      
  Cavebot.Stop()        
  Self.CloseContainers()        
  wait(2000)       
  Self.OpenMainBackpack(minimize)            
  wait(ping + 1000)     
  Cavebot.Start()           
  end    
  end    
  lastTime = os.clock()
  end)
  local script = string.dump(
  function()
  minimize = true -- minimizar a mainbp?
  bps = 1 -- quantidade de bps, deixar em 8
  local lastTime = os.clock()
  local main = Container.GetFirst()
  Module.New("reconnect", function()
  if (os.difftime(os.clock(), lastTime) >= 1) then
  local ping = Self.Ping()
  while (#Container.GetAll() < bps) do  
  Cavebot.Stop()
  Self.CloseContainers()    
  wait(2000)
  Self.OpenMainBackpack(minimize)    
  wait(ping + 1000)     
  Cavebot.Start()     
  end
  end
  lastTime = os.clock()
  end)

  end
  )

  buff=""
  for v=1,string.len(script) do --Convert our string into a hex string.
  buff=buff..'\'..string.byte(script,v)
  end

  file=io.open('encrypted.txt','w') --Output our bytecode into ascii format to encrypted.txt
  file:write(buff)
  file:flush()
  file:close()
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      loadstring(code) ()
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      loadstring(code) ()

This code in numbers above, is equivalent to the code below:

Code (2)

%pre%

I have run this script to transform the code into characters (2) for the ASCII (1) code

Code (3)

%pre%

string script criptografia lua ascii

asked by anonymous 19.07.2016 / 19:04

1 answer

Removing items from a OneToMany interface by default Spring Boot error 404

score 1 · Answer 1

As you seem to know, the string in Code 1 is a precompiled version of Code 2. It contains instructions for the Lua virtual machine. You can see a listing by changing Code 1 to io.write(code) , saving the output in a luac.out file, and then running luac -l p .

It is not easy to decompile a precompiled script. You can try using LuaDec for Lua 5.1 .