港口小按位異或perl的功能，以紅寶石

Question

我有以下的Perl腳本進行按位異或在一根繩子上用六角鍵：

#!/usr/bin/perl 

$key = pack("H*","3cb37efae7f4f376ebbd76cd"); 

print "Enter string to decode: "; 
$str=<STDIN>;chomp $str; $str =~ s/\\//g; 
$dec = decode($str); 
print "Decoded string value: $dec\n"; 

sub decode{ #Sub to decode 
    @subvar=@_; 
    my $sqlstr = $subvar[0]; 
    $cipher = unpack("u", $sqlstr); 
    $plain = $cipher^$key; 
    return substr($plain, 0, length($cipher)); 
} 

輸出示例運行它：

$ perl deXOR.pl 
Enter string to decode: (?LM-D\=^5DB$ \n 
Decoded string value: Bx3k8aaW 

我試圖將它移植到Ruby，但我做錯了什麼，結果是不一樣的：

#!/usr/bin/env ruby 

key = ['3cb37efae7f4f376ebbd76cd'].pack('H*') 

print "Enter string to decode: " 
STDOUT.flush 
a_string = gets 
a_string.chomp! 
a_string = a_string.gsub(/\//, "") 
dec = String(key) 
puts "Decoded string value: "+dec 

class String 
    def xor(key) 
    text = dup 
    text.length.times {|n| text[n] = (text[n].ord^key[n.modulo key.size].ord).chr } 
    text 
    end 
end 

示例輸出：

$ ruby deXOR.rb 
Enter string to decode: (?LM-D\=^5DB$ \n 
Decoded string value: <³~úçôóvë½vÍ 

我在做什麼錯？有什麼想法嗎？謝謝！

改變，依然一片狼藉......

key = ['3cb37efae7f4f376ebbd76cd'].pack('H*') 

def xor(text, key) 
    text.length.times {|n| text[n] = (text[n].ord^key[n.modulo key.size].ord).chr} 
    text 
end 

print "Enter string to decode: " 
STDOUT.flush 
a_string = gets 
a_string.chomp! 
a_string = a_string.gsub(/\//, "") 
dec = xor(a_string, key) 
puts "Decoded string value: "+dec 

輸出：

$ ruby deXOR.rb 
Enter string to decode: (?LM-D\=^5DB$ \n 
Decoded string value: 2·Ê°¯Kµ2" 

工作版本Digitaka的幫助：

key = ['3cb37efae7f4f376ebbd76cd'].pack('H*') 

def decode(str, key) 
    text = str.dup 
    text.length.times { |n| text[n] = (text[n].ord^key[n.modulo key.size].ord).chr } 
    text 
end 

print "Enter string to decode: " 
STDOUT.flush 
a_string = gets 
a_string.chomp! 
a_string = a_string.gsub(/\\n/, "") 
a_string = a_string.gsub(/\\/, "") 
a_string = a_string.unpack('u')[0] 
dec = decode(a_string,key) 
puts "Decoded string value: "+dec 

輸出：

$ ruby deXOR.rb 
Enter string to decode: (?LM-D=^5DB$ \n 
Decoded string value: Bx3k8aaW 

Answer 1

在perl中，您的代碼是uudecodeing輸入的字符串，並且相當於ruby中沒有發生。這個片斷uudecodes和解碼像Perl代碼：

key = ['3cb37efae7f4f376ebbd76cd'].pack('H*') 

# took some liberties to simplify the input text code 
istr = "(?LM-D=^5DB$ ".unpack('u')[0] 

def decode(str, key) 
    text = str.dup 
    text.length.times { |n| text[n] = (text[n].ord^key[n.modulo key.size].ord).chr } 
    text 
end 

puts decode(istr,key) 
# => Bx3k8aaW