66 lines
1.5 KiB
Lua
66 lines
1.5 KiB
Lua
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--与 同为1，则为1
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--或 有一个为1，则为1
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--非 true为 false，其余为true
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--异或 相同为0，不同为1
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BitUtl = {}
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local __andBit = function(left, right) --与
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return (left == 1 and right == 1) and 1 or 0
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end
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local __orBit = function(left, right) --或
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return (left == 1 or right == 1) and 1 or 0
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end
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local __xorBit = function(left, right) --异或
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return (left + right) == 1 and 1 or 0
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end
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local __base = function(left, right, op) --对每一位进行op运算，然后将值返回
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if left < right then
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left, right = right, left
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end
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local res = 0
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local shift = 1
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while left ~= 0 do
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local ra = left % 2 --取得每一位(最右边)
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local rb = right % 2
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res = shift * op(ra, rb) + res
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shift = shift * 2
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left = math.modf(left / 2) --右移
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right = math.modf(right / 2)
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end
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return res
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end
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---@public 与
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function BitUtl.andOp(left, right)
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return __base(left, right, __andBit)
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end
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---@public 或
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function BitUtl.orOp(left, right)
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return __base(left, right, __orBit)
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end
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---@public 异或
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function BitUtl.xorOp(left, right)
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return __base(left, right, __xorBit)
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end
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---@public 非
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function BitUtl.notOp(left)
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return left > 0 and -(left + 1) or -left - 1
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end
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---@public left左移num位
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function BitUtl.lShiftOp(left, num)
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return left * (2 ^ num)
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end
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---@public right右移num位
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function BitUtl.rShiftOp(left, num)
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return math.floor(left / (2 ^ num))
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end
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return BitUtl
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