Lua: print integer as a binary

Question

How can I represent integer as Binary?

so I can print 7 as 111

Answer 1

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 2

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 3

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 4

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 5

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 6

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