CODEWITHSUNDEEP
luabit-manipulationoperators
Fusseldieb
Fusseldieb

Reputation: 1374

"Bitwise AND" in Lua

I'm trying to translate a code from C to Lua and I'm facing a problem. How can I translate a Bitwise AND in Lua? The source C code contains:

if ((command&0x80)==0)
   ...

How can this be done in Lua?

I am using Lua 5.1.4-8

Upvotes: 10

Views: 26917

Answers (5)

Egor Skriptunoff
Egor Skriptunoff

Reputation: 23727

Implementation of bitwise operations in Lua 5.1 for non-negative 32-bit integers

OR, XOR, AND = 1, 3, 4

function bitoper(a, b, oper)
   local r, m, s = 0, 2^31
   repeat
      s,a,b = a+b+m, a%m, b%m
      r,m = r + m*oper%(s-a-b), m/2
   until m < 1
   return r
end

print(bitoper(6,3,OR))   --> 7
print(bitoper(6,3,XOR))  --> 5
print(bitoper(6,3,AND))  --> 2

P.S.
Lua 5.1 also has non-standard (external) library written in C: require"bit"; x=bit.band(y,z)
Lua 5.2 has standard bit32 library for bitwise functions: x=bit32.band(y,z)
Lua 5.3+ has int64 datatype with native bitwise operators: x=y&z

Upvotes: 14

Aaditya Deshpande
Aaditya Deshpande

Reputation: 41

This answer is specifically for Lua 5.1.X

you can use

if( (bit.band(command,0x80)) == 0) then
...

in Lua 5.3.X and onwards it's very straight forward...

print(5 & 6)

hope that helped 😉

Upvotes: 3

C..
C..

Reputation: 11

Here's an example of how i bitwise-and a value with a constant 0x8000:

result = (value % 65536) - (value % 32768)  -- bitwise and 0x8000

Upvotes: 1

Vurdalakov
Vurdalakov

Reputation: 109

In case you use Adobe Lightroom Lua, Lightroom SDK contains LrMath.bitAnd() method for "bitwise AND" operation:

-- x = a AND b

local a = 11
local b = 6
local x  = import 'LrMath'.bitAnd(a, b)

-- x is 2

And there are also LrMath.bitOr(a, b) and LrMath.bitXor(a, b) methods for "bitwise OR" and "biwise XOR" operations.

Upvotes: 0

ryanpattison
ryanpattison

Reputation: 6251

Here is a basic, isolated bitwise-and implementation in pure Lua 5.1:

function bitand(a, b)
    local result = 0
    local bitval = 1
    while a > 0 and b > 0 do
      if a % 2 == 1 and b % 2 == 1 then -- test the rightmost bits
          result = result + bitval      -- set the current bit
      end
      bitval = bitval * 2 -- shift left
      a = math.floor(a/2) -- shift right
      b = math.floor(b/2)
    end
    return result
end

usage:

print(bitand(tonumber("1101", 2), tonumber("1001", 2))) -- prints 9 (1001)

Upvotes: 5

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