有按位算子定律吗? [英] Are there any Bitwise Operator Laws?

问题描述

从代数定律的角度考虑，我想知道在位操作领域中是否存在与代数类似的任何官方指南.

Thinking in terms of Algebraic laws, I was wondering if there are any official guide lines which exist in the realm of bit manipulations, similar to Algebra.

代数示例

a-b =/= b-a

让 a = 7 和 b = 5

• a-b = 2
• b-a = -2

让 a = 10 和 b = 3

• a-b = 7
• b-a = -7

因此如果>b ， b-a 等于 a-b 的负值.因此，我们可以说

| a-b |= | b-a | .

Thus if a > b, b - a will be the negative equivalent to a - b. Because of this, we can say

|a - b| = |b - a|.

其中 | x | 表示 x 的绝对值.

按位示例

a |b =/= a + b

      00001010 = 10
  OR  00000101 = 5 
  -----------------
      00001111 = 15

请注意无符号字节操作: 10 |5 = 15 ，与 10 + 5 = 15

Note the unsigned byte manipulation: 10 | 5 = 15, which is synonymous with 10 + 5 = 15

但是，如果 a 和 b 都等于5并且我们对它们进行 OR ，则结果将为5，因为 a =b ，这意味着我们只是将相同的确切位彼此进行比较，因此不会产生新的结果.

However, if both a and b equal 5 and we OR them, the result would be 5, because a = b, which means we're just comparing the same exact bits with each other, thus resulting in nothing new.

同样，如果 b = 7 ， a = 10 并且我们对它们进行 OR ，我们将有15.这是因为

Likewise, if b = 7, a = 10 and we OR them we'll have 15. This is because

    00001010 = 10
 OR 00000111 = 7
 -----------------
    00001111 = 15

因此，我们可以有效地得出以下结论:b =/= a + b .

So, we can effectively conclude that a | b =/= a + b.

推荐答案

按位运算只是在操作数的相应位之间应用的布尔运算符，其遵循类似于布尔代数定律的定律，例如:

Bitwise operations that are just a boolean operator applied between corresponding bits of the operands follow laws analogous to the laws of Boolean algebra, for example:

• AND(&):可交换，关联，身份(0xFF)，An灭(0x00)，等幂
• OR(|):可交换，关联，身份(0x00)、,灭(0xFF)，等幂
• XOR(^):可交换，关联，身份(0x00)，反(自身)
• NOT(〜):反(本身)

• AND (&) : Commutative, Associative, Identity (0xFF), Annihilator (0x00), Idempotent
• OR (|) : Commutative, Associative, Identity (0x00), Annihilator (0xFF), Idempotent
• XOR (^) : Commutative, Associative, Identity (0x00), Inverse (itself)
• NOT (~) : Inverse (itself)

AND和OR相互吸收:

AND and OR absorb each other:

• a&(a | b)= a
• a |(a& b)= a

有一些分布式运算符，例如:

There are some pairs of distributive operators, such as:

• AND或: a&(b | c)=(a& b)|(a& c)
• 并通过XOR: a&(b ^ c)=(a& b)^(a& c)
• 或与AND: a |(b& c)=(a | b)&(a | c)

但是请注意，XOR不会通过AND或OR进行分配，OR也不会通过XOR进行分配.

Note however that XOR does not distribute over AND or OR, and neither does OR distribute over XOR.

DeMorgans法以各种形式适用:

DeMorgans law applies in its various forms:

• 〜(a& b)=〜a |〜b
• 〜(a | b)=〜a&〜b

可以通过对字段ℤ/2ℤ进行推理来找到一些与XOR和AND相关的定律，其中加法对应于XOR且与AND相乘:

Some laws that relate XOR and AND can be found by reasoning about the field ℤ/2ℤ, in which addition corresponds to XOR and multiplication to AND:

• AND通过XOR分发
• 计算和的乘积:(a ^ b)&(c ^ d)=(a& c)^(a& d)^(b& c)^(b& d)

有些定律将算术和按位运算相结合:

There are some laws that combine arithmetic and bitwise operations:

• 添加以下内容减去: a-b =〜(〜a + b)
• 添加另存为: a + b =(a ^ b)+((a& b)<<< 1)
• 将 min 转换为 max ，反之亦然: min(a，b)=〜max(〜a，〜b)， max(a，b)=〜min(〜a，〜b)

• Subtract by adding: a - b = ~(~a + b)
• Add carries separately: a + b = (a ^ b) + ((a & b) << 1)
• Turn min into max and vice versa: min(a, b) = ~max(~a, ~b), max(a, b) = ~min(~a, ~b)

由于位被破坏"，移位没有反作用力

Shifts have no inverse because of the "destruction" of bits pushed to the edge

左移(<<):关联，分配，标识(0x00)

left shift (<<) : Associative, Distributive, Identity (0x00)

右移(>>):关联，分配，标识(0x00)

right shift (>>) : Associative, Distributive, Identity (0x00)

向左旋转(rl):关联，分配，标识(0x00)，逆向( rr )

rotate left (rl) : Associative, Distributive, Identity (0x00), Inverse (rr)

向右旋转(rr):关联，分配，标识(0x00)，逆向( rl )

rotate right (rr) : Associative, Distributive, Identity (0x00), Inverse (rl)

虽然平移没有反函数，但某些包含平移的表达式却由于其他定律而具有反函数，例如:

While shifts have no inverse, some expressions involving shifts do have inverses as consequence of other laws, for example:

• x +(x<< k)具有反函数，因为它实际上是乘以奇数的乘积，并且奇数具有模幂为2的模乘逆.对于 x +(x<< 1)= x * 3 ，反函数为 x * 0xAAAAAAAB (对于32位，请调整其他大小的常数)
• x ^(x<< k)出于相似的原因而具有逆，但是通过与无进位乘法的对应来实现.
• 类似地， x ^(x>> k)(无符号右移)具有反函数，它只是上面的镜像".

• x + (x << k) has an inverse, because it is effectively a multiplication by an odd number and odd numbers have an modular multiplicative inverse modulo a power of two. For x + (x << 1) = x * 3, the inverse is x * 0xAAAAAAAB (for 32 bits, adjust the constant for other sizes)
• x ^ (x << k) has an inverse for a similar reason, but through the correspondence with carryless multiplication.
• Similarly x ^ (x >> k) (with unsigned right shift) has an inverse, it's just the "mirror image" of the above.

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