解码base45字符串
解码base45字符串
提问于 2021-06-24 19:14:59
我们正在尝试实现新的欧盟冠状病毒测试/疫苗接种证书的验证，但无法使base45解码工作。
规范在这里：https://datatracker.ietf.org/doc/draft-faltstrom-base45/
我们几乎完成了我们的课程，但有时我们会得到错误的值。
目标是：
Encoding example 1: The string "AB" is the byte sequence [65 66].
The 16 bit value is 65 * 256 + 66 = 16706. 16706 equals 11 + 45 * 11
+ 45 * 45 * 8 so the sequence in base 45 is [11 11 8]. By looking up
these values in the Table 1 we get the encoded string "BB8".
Encoding example 2: The string "Hello!!" as ASCII is the byte
sequence [72 101 108 108 111 33 33]. If we look at each 16 bit
value, it is [18533 27756 28449 33]. Note the 33 for the last byte.
When looking at the values modulo 45, we get [[38 6 9] [36 31 13] [9
2 14] [33 0]] where the last byte is represented by two. By looking
up these values in the Table 1 we get the encoded string "%69
VD92EX0".
Encoding example 3: The string "base-45" as ASCII is the byte
sequence [98 97 115 101 45 52 53]. If we look at each 16 bit value,
it is [25185 29541 11572 53]. Note the 53 for the last byte. When
looking at the values modulo 45, we get [[30 19 12] [21 26 14] [7 32
5] [8 1]] where the last byte is represented by two. By looking up
these values in the Table 1 we get the encoded string "UJCLQE7W581".下面是我的当前代码，它产生了错误的值：
class Base45
ALPHABET = {
"00" => "0",
"01" => "1",
"02" => "2",
"03" => "3",
"04" => "4",
"05" => "5",
"06" => "6",
"07" => "7",
"08" => "8",
"09" => "9",
"10" => "A",
"11" => "B",
"12" => "C",
"13" => "D",
"14" => "E",
"15" => "F",
"16" => "G",
"17" => "H",
"18" => "I",
"19" => "J",
"20" => "K",
"21" => "L",
"22" => "M",
"23" => "N",
"24" => "O",
"25" => "P",
"26" => "Q",
"27" => "R",
"28" => "S",
"29" => "T",
"30" => "U",
"31" => "V",
"32" => "W",
"33" => "X",
"34" => "Y",
"35" => "Z",
"36" => " ",
"37" => "$",
"38" => "%",
"39" => "*",
"40" => "+",
"41" => "-",
"42" => ".",
"43" => "/",
"44" => ":"
}.freeze
def self.encode_base45(text)
restsumme = text.unpack('S>*')
# not sure what this is doing, but without it, it works worse :D
restsumme << text.bytes[-1] if text.bytes.size > 2 && text.bytes[-1] < 256
bytearr = restsumme.map do |bytes|
arr = []
multiplier, rest = bytes.divmod(45**2)
arr << multiplier if multiplier > 0
multiplier, rest = rest.divmod(45)
arr << multiplier if multiplier > 0
arr << rest if rest > 0
arr.reverse
end
return bytearr.flatten.map{|a| ALPHABET[a.to_s.rjust(2, "0")]}.join
end
def self.decode_base45(text)
arr = text.split("").map do |char|
ALPHABET.invert[char]
end
textarr = arr.each_slice(3).to_a.map do |group|
subarr = group.map.with_index do |val, index|
val.to_i * (45**index)
end
ap subarr
subarr.sum
end
return textarr.pack("S>*") # returns wrong values
end
end结果：
Base45.encode_base45("AB")
=> "BB8" # works
Base45.decode_base45("BB8")
=> "AB" # works
Base45.encode_base45("Hello!!")
=> "%69 VD92EX" # works
Base45.decode_base45("BB8")
=> "Hello!\x00!" # wrong \x00
Base45.encode_base45("base-45")
=> "UJCLQE7W581" # works
Base45.decode_base45("UJCLQE7W581")
=> "base-4\x005" # wrong \x00任何建议都值得感谢:(
Stack Overflow用户
发布于 2021-06-24 19:59:52
如果你想要一种简单的方式：
return textarr.map{|x| x<256 ? [x].pack("C*") : [x].pack("n*") }.join看看这个方案，这感觉像是一种奇怪的编码方式，因为我们正在处理数字...如果是我，我会从字符串的尾部开始，直到字符串的头部，但那是因为我们使用的是数字。
无论如何，我的bodge之所以有效，是因为它将小元素/数字视为8位无符号，而不是16位无符号。
..。
稍微美观一点，但可能也好不到哪里去：
def self.decode_base45(text)
arr = text.split("").map do |char|
ALPHABET.invert[char]
end
textarr = arr.each_slice(3).to_a.map do |group|
subarr = group.map.with_index do |val, index|
val.to_i * (45**index)
end
ap subarr
subarr.sum.divmod(256)
end.flatten.reject(&:zero?)
return textarr.pack("C*") # returns wrong values
end页面原文内容由Stack Overflow提供。腾讯云小微IT领域专用引擎提供翻译支持
原文链接：
https://stackoverflow.com/questions/68114693
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