Interview question: Explanation of the implementation idea of ​​reversing binary bits (PHP general version)

This is an interview question, and some students expressed that they could not understand it. It is not difficult to give you a simple training, but to write a perfect comparison tests the basic skills, and also requires some logical thinking ability. Since the students' direction is PHP, then I will use PHP to explain it. At the same time, it also tells everyone that learning PHP is not It is said that as long as you can write two sentences of echo "hello world", or loop output to a web page, you can call PHP.

Here are a few pieces of knowledge: (Recommended learning: PHP video tutorial)

##1 , a number occupies one byte, that is, 8 bits

echo bindec("00000001");   //bindec函数可以让你体会到 二进制和10进制之间的 骚转换<br/>

#2 in PHP. Add decimal numbers 1 1 =2 (This tip is very important~~, Carefully understand)

Answer: 00000010 This is exactly 2 (2 raised to the power of 1)

Answer: 00000011 This is exactly 3 (2 to the power of 1 2 to the power of 0 = 2 1 = 3)

Answer: 00000100 This is 4 (2 to the power of 2)

Answer: 00000101 This guy is 4 (2 to the power of 2, 2 to the power of 0 = 4 1 = 5)

Start to solve the problem

1. First, there must be 2 variables,

2. 3 steps

Answer: Just perform a logical AND operation on the original value and 1 (that is, 00000001) (1&1 is 1, 1&0 or 0&1 are always 0)

3) Next, move $shit to the right by 1 position

The complete code is as follows

function rev($n)<br/>{<br/>    $xxoo = 0;<br/>    for ($i = 0; $i < 8; $i++) {<br/>        $xxoo = $xxoo << 1;<br/>        if (($n & 1) == 1) {<br/>            $xxoo++;<br/>        }<br/>        $n = $n >> 1;<br/>    }<br/>    return $xxoo;<br/>}<br/>echo decbin(rev(5));<br/>

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But be careful The thing is, the above function supports 1-byte numbers (only supports 8 digits)

general digits (this code is not available online~~~). Let’s think about it and understand it, so I won’t explain too much. You need to have some PHP code skills:

function rev($n)<br/>{<br/>    $num=intval(strlen(decbin($n))/8); //整除 8<br/>    if($num==0)<br/>        $bitLen=8;//最小8位<br/>    else<br/>    {<br/>        if((strlen(decbin($n)) % 8)>0)<br/>            $bitLen=($num+1)*8;<br/>        else<br/>            $bitLen=$num*8;<br/>    }<br/>    echo “原始值二进制:”.str_pad(decbin($n),$bitLen,’0′,STR_PAD_LEFT).”<br/>”;<br/>    $xxoo = 0;<br/>    for ($i = 0; $i < $bitLen; $i++) {<br/>        $xxoo = $xxoo << 1;<br/>        if (($n & 1) == 1) {<br/>            $xxoo++;<br/>        }<br/>        $n = $n >> 1;<br/>    }<br/>    echo “反转后值二进制:”.str_pad(decbin($xxoo),$bitLen,’0′,STR_PAD_LEFT).”<br/>”;<br/>    return $xxoo;<br/>}<br/>

Call test

echo rev(4);<br/>echo rev(43261596);<br/>

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Result

原始值二进制:00000100<br/>反转后值二进制:00100000<br/>32原始值二进制:00000010100101000001111010011100<br/>反转后值二进制:00111001011110000010100101000000<br/>964176192<br/>

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