This modification to ComputerCraft locks Channels 655** so they require two passwords to open them for listening. It also uses password hashing to keep your passwords secret
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Posted 09 August 2013 - 04:47 PM
Posted 09 August 2013 - 05:22 PM
Posted 09 August 2013 - 05:23 PM
Posted 09 August 2013 - 05:55 PM
MysticT, on 09 August 2013 - 05:22 PM, said:
if string.find(s, "fs%.")==nil then error("don't try and change fs to somthing else") end
Posted 09 August 2013 - 06:00 PM
Posted 09 August 2013 - 06:04 PM
Elrond1369, on 09 August 2013 - 05:55 PM, said:
PixelToast, on 09 August 2013 - 05:23 PM, said:
Posted 09 August 2013 - 06:31 PM
MysticT, on 09 August 2013 - 06:04 PM, said:
Elrond1369, on 09 August 2013 - 05:55 PM, said:
PixelToast, on 09 August 2013 - 05:23 PM, said:
fs = table.remove(fs, 14)
Posted 20 August 2013 - 09:16 PM
function hash(pass1, pass2) local key = 32 local x local y local pass local output while key % 32 == 0 do key = math.random(10000, 99999) end x = string.len(pass1) y = 1 output = 0 while y ~= x+1 do output = output + string.byte(string.sub(pass1, y, y+1)) output = output * 2 y = y + 1 end pass1 = output x = string.len(pass2) y = 1 output = 0 while y ~= x+1 do output = output + string.byte(string.sub(pass2, y, y+1)) output = output * 2 y = y + 1 end pass2 = output pass = rightRotate(pass1, key) pass = leftRotate(pass, pass2) return pass endUhm
Posted 20 August 2013 - 09:26 PM
Posted 21 August 2013 - 08:57 PM
NeverCast, on 20 August 2013 - 09:26 PM, said:
Posted 22 August 2013 - 09:39 PM
key=math.random(10000, 99999)you know, you could just multiply a random number by 32 and it will automatically become %32
x = string.len(pass2) y = 1 output = 0 while y ~= x+1 do output = output + string.byte(string.sub(pass2, y, y+1)) output = output * 2 y = y + 1 endthis.
Posted 26 August 2013 - 04:46 PM
PixelToast, on 22 August 2013 - 09:39 PM, said:
key=math.random(10000, 99999)you know, you could just multiply a random number by 32 and it will automatically become %32
x = string.len(pass2) y = 1 output = 0 while y ~= x+1 do output = output + string.byte(string.sub(pass2, y, y+1)) output = output * 2 y = y + 1 endthis.
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