一个关于PHP十进制到二进制的算法

Question

$a = 1; 
$b = 2； 
$c = 4; 
$d = 8; 
$e = 16; 
$f = 32; 
$g = 64; 
    . 
    . 
    . 

上面的序列是n的2次幂，$ n是上面几个序列的数目，如果给你$ n，用一个算法找到$ n是由几个一起去它

Answer 1

你可以得到单个位（为你变量$ a，$ b，...）与bitwise operators。

例如： 检查该位被设置

<?php 
$n = 21; //number received from somewhere 

if ($n & 1 == 1) { 
    echo "least significant bit is set"; 
} 

if ($n & 2 == 2) { 
    echo "second least significant bit is set"; 
} 

if ($n & 5 == 5) { 
    echo "third least and least significant bits are set"; 
} 

if ($n & 3 == 1) { 
    echo "least significant bit is set and second least significant bit is unset"; 
} 
?> 

例2：按位加法和乘法

<?php 
$n1 = 1 | 8 | 16; // 1 + 8 + 16 = 25 
$n2 = 2 | 8; // 2 + 8 = 10 

echo "$n1 and $n2\n"; // output: "25 and 10" 
echo ($n1 | $n2) . "\n"; // bitwise addition 25 + 10, output: "27" 
echo ($n1 & $n2) . "\n"; // bitwise multiplication 25 * 10, output: "8" 
?> 

示例3：这就是你需要

POW（2，$ I）在这种情况下产生编号为1,2,4,8,16，...，这些编号的二进制表示是：0000001，00000010,00000100,00001000，...，

按位与操作者进行零位，其中至少一个操作数具有零位，所以你可以很容易地位得到整数位

这是如何按位和作品：1101 & 0100 = 0100，1101 & 0010 = 0000

<?php 
// get number from somewhere 
$x = 27; // binary representation 00011011 

// let's define maximum exponent of 2^$a (number of bits) 
$a = 8; // 8 bit number, so it can be 0 - 255 

$result = []; 
$resIndex = 0; 

for ($i = 0; $i <= $a; $i++) { 
    // here is the heart of algorithm - it has cancer, but it should work 
    // by that cancer I mean calling three times pow isn't effective and I see other possible optimalisation, but I let it on you 
    if ((pow(2, $i) & $x) > 0) { 
     echo pow(2, $i) . "\n"; // prints "1", "2", "8", "16" 
     $result[$resIndex] = pow(2, $i); // save it to array for later use 
    } 
}