How do I use the bitwise operator XOR in Lua?

2020-02-25 08:33发布

How can I implement bitwise operators in Lua language?
Specifically, I need a XOR operator/method.

6条回答
三岁会撩人
2楼-- · 2020-02-25 09:06



This is very simple. use NAND logic. https://en.wikipedia.org/wiki/NAND_logic

function xor(a,b)
    return not( not( a and not( a and b ) ) and not( b and not( a and b ) ) )
end

if you also need 1,0 inputs insert the following to the function

    a = a==1 or a == true   -- to accept nil, 1, 0, true or false
    b = b==1 or b == true   -- to accept nil, 1, 0, true or false

Hope this helps someone.

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兄弟一词,经得起流年.
3楼-- · 2020-02-25 09:14

Since you're referencing the floor function 3 times, using an excessive number of loops for most operations (numbers less than 2^31 don't need all 31 loops), are using the ^ operator, and aren't capitalizing on the fact that a and b might be wildly different numbers with different magnitudes, you're losing a lot of efficiency. The function also isn't localized, and you're doing two more division operations than you need to. I wrote this to be reasonably fast.

In general, you're going to see improvements of about 3 to 20 times.

local function BitXOR(a,b)--Bitwise xor
    local p,c=1,0
    while a>0 and b>0 do
        local ra,rb=a%2,b%2
        if ra~=rb then c=c+p end
        a,b,p=(a-ra)/2,(b-rb)/2,p*2
    end
    if a<b then a=b end
    while a>0 do
        local ra=a%2
        if ra>0 then c=c+p end
        a,p=(a-ra)/2,p*2
    end
    return c
end

If you need more than this, say AND, OR, and NOT, then I've got you covered there, too.

local function BitOR(a,b)--Bitwise or
    local p,c=1,0
    while a+b>0 do
        local ra,rb=a%2,b%2
        if ra+rb>0 then c=c+p end
        a,b,p=(a-ra)/2,(b-rb)/2,p*2
    end
    return c
end

local function BitNOT(n)
    local p,c=1,0
    while n>0 do
        local r=n%2
        if r<1 then c=c+p end
        n,p=(n-r)/2,p*2
    end
    return c
end

local function BitAND(a,b)--Bitwise and
    local p,c=1,0
    while a>0 and b>0 do
        local ra,rb=a%2,b%2
        if ra+rb>1 then c=c+p end
        a,b,p=(a-ra)/2,(b-rb)/2,p*2
    end
    return c
end

Don't worry, you won't need to change anything.

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冷血范
4楼-- · 2020-02-25 09:21

In Lua 5.2, you can use the bit32.bxor function.

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够拽才男人
5楼-- · 2020-02-25 09:21

From the OP; moved from question into this answer.


This is how I implemented XOR in Lua:

local floor = math.floor
function bxor (a,b)
  local r = 0
  for i = 0, 31 do
    local x = a / 2 + b / 2
    if x ~= floor (x) then
      r = r + 2^i
    end
    a = floor (a / 2)
    b = floor (b / 2)
  end
  return r
end
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女痞
6楼-- · 2020-02-25 09:25

In Lua 5.2, you can use functions in bit32 library.

In Lua 5.3, bit32 library is obsoleted because there are now native bitwise operators.

print(3 & 5)  -- bitwise and
print(3 | 5)  -- bitwise or
print(3 ~ 5)  -- bitwise xor
print(7 >> 1) -- bitwise right shift
print(7 << 1) -- bitwise left shift
print(~7)     -- bitwise not

Output:

1
7
6
3
14
-8
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SAY GOODBYE
7楼-- · 2020-02-25 09:28

If you're needing an efficient way to do bitwise shifts, I wrote an article about that a while ago. Here's some functions which wrap the technique:

function lshift(x, by)
  return x * 2 ^ by
end

function rshift(x, by)
  return math.floor(x / 2 ^ by)
end
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