Question

Perl中的程序加密很好，但解密完全出现在0。为什么会这样？我使用相同的方法来加密/解密字符串。

$string = "This is an encryption program";        # any length
$keystr = "948D5E5";                              # length <= 8

print "original  : ", $string, "\n";

$enc = hexcrypt($string, $keystr);
print "encrypted : ", $enc, "\n";

$dec = hexcrypt($enc, $keystr);
print "decrypted : ", $dec, "\n";

sub hexcrypt {
  my (@pac, $pacstr, $l, $pac, $pacenc, $format, $pacencstr);

  my @string    = split(//, @_[0]);               # the string to encrypt
  my $keystr    = @_[1];                          # the key
  my $key       = hex($keystr);                   # the key in binary
  my $keylength = length($keystr);
  my $encstr    = "";                             # accumulator for returned string

  do {
    @pac       = splice(@string, 0, $keylength);  # take out a packet of digits (or less at end)
    $pacstr    = join("", @pac);                  # into a string
    $l         = length($pacstr);
    $pac       = hex($pacstr);                    # convert to a 32-bit integer
    $pacenc    = $pac ^ ($key >> ($keylength * 4 - 4 * $l)); # shift key right if $l < $keylength, and   XOR
    $format    = "%0" . sprintf("%.d", $l) . "X"; # we want the leading zeros
    $pacencstr = sprintf($format, $pacenc);       # back into a string

    $encstr   .= $pacencstr;                      # add to encrypted string
  } while scalar @string > 0;

  return $encstr;
}

Answer 1

您的变量$pacstr是要编码的字符串的一部分，其长度等于或小于密钥的长度。

此处，您的密钥为948D5E5，长度为七个字符，因此$pacstr的第一个值是$string的前七个字符，即This is。< / p>

所以你的陈述

$pac = hex($pacstr)

正在做

$pax = hex('This is')

返回零，因为字符串中没有有效的十六进制数字。

当然，您必须在每个 Perl程序的开头遇到use strict和use warnings的劝诫？如果你在这里遵循了这个建议，你会看到以下

Illegal hexadecimal digit 'T' ignored at E:\Perl\source\xmls.pl line 33.
Illegal hexadecimal digit ' ' ignored at E:\Perl\source\xmls.pl line 33.
Illegal hexadecimal digit 'r' ignored at E:\Perl\source\xmls.pl line 33.
Illegal hexadecimal digit ' ' ignored at E:\Perl\source\xmls.pl line 33.
Illegal hexadecimal digit 'm' ignored at E:\Perl\source\xmls.pl line 33.

这是该问题的大线索。

我在Perl中的解密是全部0

1 个答案: