为什么异或结果不同，0变为1 [英] Why xor results are different, 0 becomes 1

问题描述

我想制作一个如下图所示的程序

i want to make a program like the following picture

这是我的代码

<?php

 $iv = 0;
 $Kunci = "U";

 $key =  dechex(ord($Kunci));
 $k =  sprintf("%08d",decbin(hexdec($key)));

 $c0 = sprintf("%08d", decbin($iv));
 $Cip= "0C52CCD7EDB3";
 $Cbs = array();
 $Cbs[0]= $c0;

 $Plaintext = array();
 $Cas = array();
 $P = array();  
 $m= 1;
 $n=1;

//$Cbs= 
$Csplit = str_split($Cip, 2);
$Cas= str_split($Cip,2);

        for ($i=0; $i<count($Csplit); $i++) { 

            $Cbs[$m] = sprintf("%08d",decbin(hexdec($Csplit[$i])));
            $m++;

        }



        for($i=0; $i < count($Cas); $i++){
            $Cas[$i] = sprintf("%08d",decbin(hexdec($Cas[$i])));
            $Cas[$i]=bindec($Cas[$i])>>1;
            if($Cas[$i] % 2 <> 0)$Cas[$i]+=128;
            $Cas[$i]=sprintf("%08d", decbin($Cas[$i]));

        }


 foreach($Cas as $cas_item) {
$prev_c = $Cbs[$n-1];    

$P[$n] = _xor($cas_item, $k);

 $P[$n] = _xor($P[$n], $prev_c);


$Plaintext[$n] = chr(bindec($P[$n]));

 $n++; 
 }

function _xor($text,$key){
for($i=0; $i<strlen($text); $i++){
  $text[$i] = intval($text[$i])^intval($key[$i]);

}
  return $text;
}

print_r($Csplit);
echo "<br/>";
print_r($Cbs);
echo "<br/>";
print_r($Cas);
echo "<br/>";
print_r($P);
echo "<br/>";
print_r($Plaintext);

?>

Cbs =移位仓之前 CAS =移位仓后 这样一来，程序代码就可以工作了，但是数组2和数组5是错误的.前面的二进制位代码应为0，而不是1. 输出 :

Cbs = before shift biner Cas = after shift biner and this comes out, the program code works but array 2 and array 5 are wrong. the binary bit code in front should be 0, not 1. Output :

数组2应该是01110000而不是11110000，数组5应该是01110100但结果是11110100.为什么前面的0是1?

array 2 should be 01110000 instead of 11110000, and array 5 should be 01110100 but result is 11110100. why is 0 in front being 1?

推荐答案

右移时，请注意有符号和无符号移位的区别. (也称为算术或逻辑移位)

When shifting right, beware of the difference of signed and unsigned shift. (also called arithmetic or logical shift)

按符号右移的8位值11101000将为11110100.

8-bit value 11101000 right shifted as signed will be 11110100.

要点是，如果您要向右移动一个有符号的值，那么最高的位将被复制到移入的新位中.如果您要转移无符号的值，那么最高的位将被移为零.

The point is if you are shifting a signed value to the right, the uppermost bit is duplicated into the new bits moving in. If you are shifting unsigned values, the uppermost bits move in zeroes.

缺少无符号整数数据类型的语言还有另一个右移运算符>>>，表示要进行无符号(或逻辑")移位.在PHP和Java中就是这种情况.

Languages that lack unsigned integer datatypes have another right-shift operator >>> to indicate that an unsigned (or 'logical') shift is meant. This is the case in PHP and in Java.

这仅适用于右移.永不离开.关键是右移将导致行为被二分频.

This only applies to right-shifts. Never to left. The point is that a right-shift will result in a divide-by-two behaviour.

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