Unreal Software - Thema: Bitwise operations

Bitwise operations

Offline

26.04.11 01:18:40

I wrote this so I could do bit operations in Lua. I thought someone else might need it so I am posting it.

Function list

Zitat

tobits (n)
     n - number to convert to binary/bits

     returns a table of bits

BitOp (n, n2, bits, c)
     n - first number
     n2 second number
     bits - max bits (default 8 (1 byte/octet))
     c - bitop function to use (default and)

     performs c on each bit in n and n2

NOT (n)
     n - number

     returns n except with the bits negated

LSHIFT (n, s)
     n - number
     s - number of shifts (default 1)

     shifts each bit in n to the left s times

RSHIFT (n, s)
     n - number
     s - number of shifts (default 1)

     shifts each bit in n to the right s times

BitOp functions
Use these as the c parameter of BitOp

Zitat

bitop_or perform or on the bits
bitop_xor perform exclusive or on the bits
bitop_and perform and on the bits

Source

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function tobits(n)
	local l = math.floor(math.log(n)/math.log(2))
	r = {}

	for i=l,0,-1 do
		local b = math.pow(2,i)
		if b <= n then
			n = n - b
			r[#r+1] = 1
		else
			r[#r+1] = 0
		end
	end
	return r
end

function BitOp(n, n2, bits, c)
	bits = bits or 8
	c = c or function(a, b) return a == 1 and b == 1 end
	n = n%2^bits
	n2 = n2%2^bits
	n = tobits(n)
	n2 = tobits(n2)
	local s = {}

	while ( #n < bits ) do
		table.insert(n, 1, 0)
	end

	while ( #n2 < bits ) do
		table.insert(n2, 1, 0)
	end

	for i = 1, bits do
		if ( c(n[i], n2[i]) ) then
			s[#s+1] = 1
		else
			s[#s+1] = 0
		end
	end

	local p = 0
	local n = 0
	for i = bits, 1, -1 do
		n = n + (s[i] * 2^p)
		p = p + 1
	end

	return n
end

bitop_xor = function(a, b)
	if ( a == 1 and b == 0 ) then
		return true
	elseif ( a == 0 and b == 1 ) then
		return true
	end

	return false
end

bitop_or = function(a, b)
	if ( a == 1 or b == 1 ) then
		return true
	end

	return false
end

bitop_and = function(a, b)
	return a == 1 and b == 1
end

function NOT(n)
	n = tobits(n)
	for i = 1, #n do
		n[i] = n[i] == 1 and 0 or 1
	end
	
	local num = 0
	local p = 0
	for i = #n, 1, -1 do
		num = num + (n[i] * 2^p)
		p = p + 1
	end
	
	return num
end

function LSHIFT(n, s)
	s = s or 1
	n = tobits(n)
	
	for i = 1, s do
		n[#n+1] = 0
	end
	
	local num, p = 0,0
	for i = #n, 1, -1 do
		num = num + (n[i] * 2^p)
		p = p + 1
	end
	
	return num
end

function RSHIFT(n, s)
	s = s or 1
	n = tobits(n)
	
	for i = 1, s do
		if ( # n == 0 ) then
			break
		end
		
		table.remove(n, #n)
	end
	
	local num,p = 0,0
	for i = #n, 1, -1 do
		num = num + (n[i] * 2^p)
		p = p + 1
	end
	
	return num
end

1× editiert, zuletzt 26.04.11 02:07:47

Re: Bitwise operations

Flacko
User Off

Offline

26.04.11 03:14:25

Darkbyte, I think it's unnecessary to have a "LSHIFT" and "RSHIFT" function. Just make a "SHIFT" function and make it so that negative values move the bits to the left and positive values move it to the right.

Something like this:

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function bit:shift(shift)
	local start, finish, increase = 1,#self,1
	if shift < 0 then start,finish,increase = #self,1,-1 end
	for i=start,finish,increase do
		if i+shift>#self or i+shift<1 then self[i]=false else self[i]=self[i+shift] end
	end
end

But I'm not sure it will work with your code...

EDIT: I also think you should use tables with boolean values instead of numbers. If you do so, you can write functions in a much nicier-looking way:

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--Binary NOT example
function bit:bnot()
	for i=1, #self do
		self[i] = not self[i]
	end
end

1× editiert, zuletzt 26.04.11 10:57:11

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Bitwise operations