php移位运算、移位操作学习笔记
位运算应用口诀
清零取位要用与，某位置一可用或
若要取反和交换，轻轻松松用异或
移位运算
要点
1 它们都是双目运算符，两个运算分量都是整形，结果也是整形。
2 "
3 ">>"右移：右边的位被挤掉。对于左边移出的空位，如果是正数则空位补0，若为负数，可能补0或补1，这取决于所用的计算机系统。
4 ">>>"运算符，右边的位被挤掉，对于左边移出的空位一概补上0。
位运算符的应用 (源操作数s 掩码mask)
(1) 按位与-- &
1 清零特定位 (mask中特定位置0，其它位为1，s=s&mask)
2 取某数中指定位 (mask中特定位置1，其它位为0，s=s&mask)
(2) 按位或-- |
常用来将源操作数某些位置1，其它位不变。 (mask中特定位置1，其它位为0 s=s|mask)
(3) 位异或-- ^
1 使特定位的值取反 (mask中特定位置1，其它位为0 s=s^mask)
2 不引入第三变量，交换两个变量的值 (设 a=a1,b=b1)
目 标 操 作 操作后状态
a=a1^b1 a=a^b a=a1^b1,b=b1
b=a1^b1^b1 b=a^b a=a1^b1,b=a1
a=b1^a1^a1 a=a^b a=b1,b=a1
二进制补码运算公式：
-x = ~x + 1 = ~(x-1)
~x = -x-1
-(~x) = x+1
~(-x) = x-1
x+y = x - ~y - 1 = (x|y)+(x&y)
x-y = x + ~y + 1 = (x|~y)-(~x&y)
x^y = (x|y)-(x&y)
x|y = (x&~y)+y
x&y = (~x|y)-~x
x==y: ~(x-y|y-x)
x!=y: x-y|y-x
x
x
x
x
应用举例
(1) 判断int型变量a是奇数还是偶数
a&1 = 0 偶数
a&1 = 1 奇数
(2) 取int型变量a的第k位 (k=0,1,2……sizeof(int))，即a>>k&1
(3) 将int型变量a的第k位清0，即a=a&~(1
(4) 将int型变量a的第k位置1， 即a=a|(1
(5) int型变量循环左移k次，即a=a>16-k (设sizeof(int)=16)
(6) int型变量a循环右移k次，即a=a>>k|a
(7)整数的平均值
对于两个整数x,y，如果用 (x+y)/2 求平均值，会产生溢出，因为 x+y 可能会大于INT_MAX，但是我们知道它们的平均值是肯定不会溢出的，我们用如下算法：
int average(int x, int y) //返回X,Y 的平均值
{
return (x&y)+((x^y)>>1);
}(8)判断一个整数是不是2的幂,对于一个数 x >= 0，判断他是不是2的幂
boolean power2(int x)
{
return ((x&(x-1))==0)&&(x!=0)；
}(9)不用temp交换两个整数
void swap(int x , int y)
{
x ^= y;
y ^= x;
x ^= y;
}(10)计算绝对值
int abs( int x )
{
int y ;
y = x >> 31 ;
return (x^y)-y ; //or: (x+y)^y
}(11)取模运算转化成位运算 (在不产生溢出的情况下)
a % (2^n) 等价于 a & (2^n - 1)
(12)乘法运算转化成位运算 (在不产生溢出的情况下)
a * (2^n) 等价于 a<< n
(13)除法运算转化成位运算 (在不产生溢出的情况下)
a / (2^n) 等价于 a>> n 例: 12/8 == 12>>3
(14) a % 2 等价于 a & 1
(15) if (x == a) x= b;
else x= a;等价于 x= a ^ b ^ x;
(16) x 的 相反数 表示为 (~x+1)
最后补充一些关于二进制位移操作
PHP主要是设计于文本操作的，其实PHP不适合做数学运算，效率也不高，不过因为这次的项目中有个东西必须使用到二进制位移操作，在PHP上面遇到了一些麻烦。
因为PHP只有32位有符号整数，没有64位长整型，也没有无符号整数。其整型的范围是-231-1～231，超出这个范围的，将被解释为浮点数。因此，0xFFFFFFFF，直接打印，显示的是4294967295，及232：
>> 0xFFFFFFFFF
4294967295
>> gettype(0xFFFFFFFF)
'double'
而在32位有符号整型中，0xFFFFFFFF应表示-1：
>> (int)0xFFFFFFFFF
-1
而PHP不支持浮点数的二进制位移操作，如果要进行，会先转换为整型，最后的结果，也将按照整型来返回：
>> 1 << 31 -2147483648 >> 1 << 30 1073741824 >> 1 << 32 1 >> 0xFFFFFFFF >> 1 -1
同时PHP的向右位移操作，高位会填充符号位，而且PHP没有提供类似Java的>>>来强制填充0：
>> 1 << 32 1 >> 0xFFFFFFFF >> 1 -1 >> 0xFFFFFFFF >> 2 -1 >> 0xFFFFFFFF >> 3 -1 >> 0xFFFFFFFF >> 31 -1
如何解决这个问题呢，我考虑过使用BCMath数学函数库，直接处理字符串表示的整数，或者是GMP/BigInt扩展等。不过我想既然使用字符串，那么我可以字符串地彻底一些，把数字转换成32个二进制的字符串，再手工填充0，最后转换回来。
不知道哪位有更好的方法，请告诉我,另外，其实代码可以扩展为任意位2进制的位移操作，这里我没有做,PHP代码如下:
<?php
/**
* 无符号32位右移
* @param mixed $x 要进行操作的数字，如果是字符串，必须是十进制形式
* @param string $bits 右移位数
* @return mixed 结果，如果超出整型范围将返回浮点数
*/
function shr32($x, $bits) {
// 位移量超出范围的两种情况
if ($bits <= 0) {
return $x;
}
if ($bits >= 32) {
return 0;
}
//转换成代表二进制数字的字符串
$bin = decbin($x);
$l = strlen($bin);
//字符串长度超出则截取底32位，长度不够，则填充高位为0到32位
if ($l > 32) {
$bin = substr($bin, $l - 32, 32);
} elseif ($l < 32) {
$bin = str_pad($bin, 32, '0', STR_PAD_LEFT);
}
//取出要移动的位数，并在左边填充0
return bindec(str_pad(substr($bin, 0, 32 - $bits) , 32, '0', STR_PAD_LEFT));
}
/**
* 无符号32位左移
* @param mixed $x 要进行操作的数字，如果是字符串，必须是十进制形式
* @param string $bits 左移位数
* @return mixed 结果，如果超出整型范围将返回浮点数
*/
function shl32($x, $bits) {
// 位移量超出范围的两种情况
if ($bits <= 0) {
return $x;
}
if ($bits >= 32) {
return 0;
}
//转换成代表二进制数字的字符串
$bin = decbin($x);
$l = strlen($bin);
//字符串长度超出则截取底32位，长度不够，则填充高位为0到32位
if ($l > 32) {
$bin = substr($bin, $l - 32, 32);
} elseif ($l < 32) {
$bin = str_pad($bin, 32, '0', STR_PAD_LEFT);
}
//取出要移动的位数，并在右边填充0
return bindec(str_pad(substr($bin, $bits) , 32, '0', STR_PAD_RIGHT));
}
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