In my Lua script, I have a variable that stores an integer found from some function that represents a hex string. I want to flip it so that, for example, if it has a value of 0x006E57E8, it will become 0x8E75E600. How do I do it in Lua?
As an alternative to Skriptunoff's very concise & readable function, here's a function doing it using bit ops:
function reverse_nibbles(int)
res = 0
for i = 0, 7 do
res = res | (((int >> (4*i)) & 0xF) << (28 - 4*i))
end
return res
end
This won't create a garbage string (and would hopefully be faster). You might want to hand-unroll the loop.
Related
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()
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.
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?
I need a base converter function for Lua. I need to convert from base 10 to base 2,3,4,5,6,7,8,9,10,11...36 how can i to this?
In the string to number direction, the function tonumber() takes an optional second argument that specifies the base to use, which may range from 2 to 36 with the obvious meaning for digits in bases greater than 10.
In the number to string direction, this can be done slightly more efficiently than Nikolaus's answer by something like this:
local floor,insert = math.floor, table.insert
function basen(n,b)
n = floor(n)
if not b or b == 10 then return tostring(n) end
local digits = "0123456789ABCDEFGHIJKLMNOPQRSTUVWXYZ"
local t = {}
local sign = ""
if n < 0 then
sign = "-"
n = -n
end
repeat
local d = (n % b) + 1
n = floor(n / b)
insert(t, 1, digits:sub(d,d))
until n == 0
return sign .. table.concat(t,"")
end
This creates fewer garbage strings to collect by using table.concat() instead of repeated calls to the string concatenation operator ... Although it makes little practical difference for strings this small, this idiom should be learned because otherwise building a buffer in a loop with the concatenation operator will actually tend to O(n2) performance while table.concat() has been designed to do substantially better.
There is an unanswered question as to whether it is more efficient to push the digits on a stack in the table t with calls to table.insert(t,1,digit), or to append them to the end with t[#t+1]=digit, followed by a call to string.reverse() to put the digits in the right order. I'll leave the benchmarking to the student. Note that although the code I pasted here does run and appears to get correct answers, there may other opportunities to tune it further.
For example, the common case of base 10 is culled off and handled with the built in tostring() function. But similar culls can be done for bases 8 and 16 which have conversion specifiers for string.format() ("%o" and "%x", respectively).
Also, neither Nikolaus's solution nor mine handle non-integers particularly well. I emphasize that here by forcing the value n to an integer with math.floor() at the beginning.
Correctly converting a general floating point value to any base (even base 10) is fraught with subtleties, which I leave as an exercise to the reader.
you can use a loop to convert an integer into a string containting the required base. for bases below 10 use the following code, if you need a base larger than that you need to add a line that mapps the result of x % base to a character (usign an array for example)
x = 1234
r = ""
base = 8
while x > 0 do
r = "" .. (x % base ) .. r
x = math.floor(x / base)
end
print( r );