I have a 4 byte hexadecimal value that I have a script to print out, But I want to now take that value then subtract the value C8 from it 37 times and save them as different variables each time, But the problem is I don't know how to do hexadecimal calculations in lua, If anyone can link me to any documentation on how to do this then that would be much appreciated.
You can make a hexadecimal literal in Lua by prefixing it with 0x, as stated in the reference manual. I found this by googling "lua hex"; such searches usually get good results.
"Hexadecimal numbers" aren't anything special, hexadecimal is just a way to represent numbers, same as decimal or binary. You can do 1000-0xC8 and you'll get the decimal number 800.
Code to convert:
function convertHex()
local decValue = readInteger(0x123456);
hexValue = decValue
end
function hexSubtract()
for i = 1,37 do
local value = 0xC8
hexValue = hexValue - 0xC8
result = hexValue
if i == 37 then
print(result) --Prints dec value
print(string.format('%X',result)); --Prints hex value
end
end
end
Replace 0x123456 with your address, use those functions like this convertHex(),hexSubtract()
Related
I've got the following code in lua:
array = {}
for i=1, 30 do
mem = tostring(0x29A300 + (i * 0x11CC))
array[i] = "Ref Mem" .. i .. ": " .. mem
end
But when I do the hexadecimal addition to store in mem, I get a decimal value out.
Is there a way to keep the hex formatting that I used to do the calculation?
Also is there a way for me to remove the 0x in the string?
I've tried using tonumber but that gives me an error.
There hasn't been anything else I've tried as I'm completely new to lua.
I'm trying to create a list of integers, similar to python where one would say
x = input("Enter String").split() # 1 2 3 5
x = list(map(int,x)) # Converts x = "1","2",3","5" to x = 1,2,3,5
Here's my code asking for the input, then splitting the input into a table, i need help converting the contents of the table to integers as they're being referenced later in a function, and i'm getting a string vs integer comparison error. I've tried changing the split for-loop to take a number but that doesn't work, I'm familiar with a python conversion but not with Lua so I'm looking for some guidance in converting my table or handling this better.
function main()
print("Hello Welcome the to Change Maker - LUA Edition")
print("Enter a series of change denominations, separated by spaces")
input = io.read()
deno = {}
for word in input:gmatch("%w+") do table.insert(deno,word) end
end
--Would This Work?:
--for num in input:gmatch("%d+") do table.insert(deno,num) end
Just convert your number-strings to numbers using tonumber
local number = tonumber("1")
So
for num in input:gmatch("%d+") do table.insert(deno,tonumber(num)) end
Should do the trick
I am trying to return very long integer number but my result returns as
"7.6561197971049e+016".
How do I make it return 76561197971049296 ?
local id64 = 76561197960265728
Z = string.match("STEAM_0:0:5391784", 'STEAM_%d+:%d+:(%d+)')
Y = string.match("STEAM_0:0:5391784", 'STEAM_%d+:(%d+):%d+')
--For 64-bit systems
--Let X, Y and Z constants be defined by the SteamID: STEAM_X:Y:Z.
--Let V be SteamID64 identifier of the account type (0x0110000100000000 in hexadecimal format).
--Using the formula W=Z*2+V+Y
if Z == nil then
return "none"
else
return Z*2+id64+Y
end
I installed lbc arbitrary precision now with this code
return bc.add(bc.number(id64),bc.number(2)):tostring()
it returns 70000000000000002 but if I delete 3 digits from id64 it displays correctly.
How can I get correct result without deleting the digits?
You need to use strings for long numbers. Otherwise, the Lua lexer converts them to doubles and loses precision in this case. Here is code using my lbc:
local bc=require"bc"
local id64=bc.number"76561197960265728"
local Y,Z=string.match("STEAM_0:0:5391784",'STEAM_%d+:(%d+):(%d+)')
if Z == nil then
return "none"
else
return (Z*2+id64+Y):tostring()
end
check out this library for arbitrary precision arithmetics. this so post might be of interest to you as well.
Assuming your implementation of Lua supports that many significant digits in the number type, your return statement is returning that result.
You're probably seeing exponential notation when you convert the number to a string or printing it. You can use the string.format function to control the conversion:
assert( "76561197971049296" == string.format("%0.17g", 76561197971049296))
If number is an IEEE-754 double, then it doesn't work. You do have to know how your Lua is implemented and keep in mind the the technical limitations.
If you have luajit installed, you can do this:
local ffi = require("ffi")
steamid64 = tostring(ffi.new("uint64_t", 76561197960265728) + ffi.new("uint64_t", tonumber(accountid)))
steamid64 = string.sub(steamid64, 1, -4) -- to remove 'ULL at the end'
Hope it helps.
I am trying to get a random 16 digit number in Lua. What I have written isn't working out for me when logically it should. How does math.random work with exponents?
This is what I keep getting.
> return math.random(10^15, 10^16)
> -1637272360
If you want to have a 16 digit number, try generating them this way:
local fmt = "%d%07d%08d"
local random = math.random
local num = fmt:format(random(1, 9), random(0, 10^7), random(0, 10^8))
and then keep the variable num in string type. As a number, it converts the values to exponential form(because of the very large; in your case > 10^14; exponential value) or otherwise, you can store them as a(n) Hex string?
Easy question here, probably, but searching did not find a similar question.
The # operator finds the length of a string, among other things, great. But with Lua being dynamically typed, thus no conversion operators, how does one type a number as a string in order to determine its length?
For example suppose I want to print the factorials from 1 to 9 in a formatted table.
i,F = 1,1
while i<10 do
print(i.."! == "..string.rep("0",10-#F)..F)
i=i+1
F=F*i
end
error: attempt to get length of global 'F' (a number value)
why not use tostring(F) to convert F to a string?
Alternatively,
length = math.floor(math.log10(number)+1)
Careful though, this will only work where n > 0!
There are probably a dozen ways to do this. The easy way is to use tostring as Dan mentions. You could also concatenate an empty string, e.g. F_str=""..F to get F_str as a string representation. But since you are trying to output a formatted string, use the string.format method to do all the hard work for you:
i,F = 1,1
while i<10 do
print(string.format("%01d! == %010d", i, F))
i=i+1
F=F*i
end
Isn't while tostring(F).len < 10 do useful?