Lua: print integer as a binary - binary

Question

How can I represent integer as Binary?
so I can print 7 as 111

Answer 1

You write a function to do this.
num=7
function toBits(num)
-- returns a table of bits, least significant first.
local t={} -- will contain the bits
while num>0 do
rest=math.fmod(num,2)
t[#t+1]=rest
num=(num-rest)/2
end
return t
end
bits=toBits(num)
print(table.concat(bits))
In Lua 5.2 you've already have bitwise functions which can help you ( bit32 )
Here is the most-significant-first version, with optional leading 0 padding to a specified number of bits:
function toBits(num,bits)
-- returns a table of bits, most significant first.
bits = bits or math.max(1, select(2, math.frexp(num)))
local t = {} -- will contain the bits
for b = bits, 1, -1 do
t[b] = math.fmod(num, 2)
num = math.floor((num - t[b]) / 2)
end
return t
end

Answer 2

There's a faster way to do this that takes advantage of string.format, which converts numbers to base 8. It's trivial to then convert base 8 to binary.
--create lookup table for octal to binary
oct2bin = {
['0'] = '000',
['1'] = '001',
['2'] = '010',
['3'] = '011',
['4'] = '100',
['5'] = '101',
['6'] = '110',
['7'] = '111'
}
function getOct2bin(a) return oct2bin[a] end
function convertBin(n)
local s = string.format('%o', n)
s = s:gsub('.', getOct2bin)
return s
end
If you want to keep them all the same size, then do
s = string.format('%.22o', n)
Which gets you 66 bits. That's two extra bits at the end, since octal works in groups of 3 bits, and 64 isn't divisible by 3. If you want 33 bits, change it to 11.
If you have the BitOp library, which is available by default in LuaJIT, then you can do this:
function convertBin(n)
local t = {}
for i = 1, 32 do
n = bit.rol(n, 1)
table.insert(t, bit.band(n, 1))
end
return table.concat(t)
end
But note this only does the first 32 bits! If your number is larger than 2^32, the result wont' be correct.

Answer 3

function bits(num)
local t={}
while num>0 do
rest=num%2
table.insert(t,1,rest)
num=(num-rest)/2
end return table.concat(t)
end
Since nobody wants to use table.insert while it's useful here

Answer 4

Here is a function inspired by the accepted answer with a correct syntax which returns a table of bits in wriiten from right to left.
num=255
bits=8
function toBits(num, bits)
-- returns a table of bits
local t={} -- will contain the bits
for b=bits,1,-1 do
rest=math.fmod(num,2)
t[b]=rest
num=(num-rest)/2
end
if num==0 then return t else return {'Not enough bits to represent this number'}end
end
bits=toBits(num, bits)
print(table.concat(bits))
>>11111111

Answer 5

function reverse(t)
local nt = {} -- new table
local size = #t + 1
for k,v in ipairs(t) do
nt[size - k] = v
end
return nt
end
function tobits(num)
local t={}
while num>0 do
rest=num%2
t[#t+1]=rest
num=(num-rest)/2
end
t = reverse(t)
return table.concat(t)
end
print(tobits(7))
# 111
print(tobits(33))
# 100001
print(tobits(20))
# 10100

Answer 6

local function tobinary( number )
local str = ""
if number == 0 then
return 0
elseif number < 0 then
number = - number
str = "-"
end
local power = 0
while true do
if 2^power > number then break end
power = power + 1
end
local dot = true
while true do
power = power - 1
if dot and power < 0 then
str = str .. "."
dot = false
end
if 2^power <= number then
number = number - 2^power
str = str .. "1"
else
str = str .. "0"
end
if number == 0 and power < 1 then break end
end
return str
end
May seem more verbose but it is actually faster than other functions that use the math library functions. Works with any number, be it positive/negative/fractional...

Answer 7

local function tobits(num, str) -- tail call
str = str or "B"
if num == 0 then return str end
return tobits(
num >> 1 , -- right shift
((num & 1)==1 and "1" or "0") .. str )
end

Answer 8

This function uses a lookup table to print a binary number extracted from a hex representation. All using string manipulation essentially. Tested in lua 5.1.
local bin_lookup = {
["0"] = "0000",
["1"] = "0001",
["2"] = "0010",
["3"] = "0011",
["4"] = "0100",
["5"] = "0101",
["6"] = "0110",
["7"] = "0111",
["8"] = "1000",
["9"] = "1001",
["A"] = "1010",
["B"] = "1011",
["C"] = "1100",
["D"] = "1101",
["E"] = "1110",
["F"] = "1111"
}
local print_binary = function(value)
local hs = string.format("%.2X", value) -- convert number to HEX
local ln, str = hs:len(), "" -- get length of string
for i = 1, ln do -- loop through each hex character
local index = hs:sub(i, i) -- each character in order
str = str .. bin_lookup[index] -- lookup a table
str = str .. " " -- add a space
end
return str
end
print(print_binary(45))
#0010 1101
print(print_binary(65000))
#1111 1101 1110 1000

Answer 9

This maybe not work in lua that has no bit32 library
function toBinary(number, bits)
local bin = {}
bits = bits - 1
while bits >= 0 do --As bit32.extract(1, 0) will return number 1 and bit32.extract(1, 1) will return number 0
--I do this in reverse order because binary should like that
table.insert(bin, bit32.extract(number, bits))
bits = bits - 1
end
return bin
end
--Expected result 00000011
print(table.concat(toBinary(3, 8)))
This need at least lua 5.2 (because the code need bit32 library)

Answer 10

As by Dave, but with filled empty bits:
local function toBits(num, bits)
-- returns a table of bits, least significant first.
local t={} -- will contain the bits
bits = bits or 8
while num>0 do
rest=math.fmod(num,2)
t[#t+1]=rest
num=math.floor((num-rest)/2)
end
for i = #t+1, bits do -- fill empty bits with 0
t[i] = 0
end
return t
end
for i = 0, 255 do
local bits = toBits(i)
print(table.concat(bits, ' '))
end
Result:
0 0 0 0 0 0 0 0
1 0 0 0 0 0 0 0
0 1 0 0 0 0 0 0
1 1 0 0 0 0 0 0
0 0 1 0 0 0 0 0
1 0 1 0 0 0 0 0
...
0 1 1 1 1 1 1 1
1 1 1 1 1 1 1 1