为什么是十六进制-> base64与base64有何不同->十六进制使用打包和解包? [英] Why is hex -> base64 so different from base64 -> hex using pack and unpack?

问题描述

我使这段代码正常工作，将其从十六进制转换为base64，反之亦然.我从另一个SO问题中得到了to_base64，然后我写了to_hex并带有一些猜测和反复试验.

I got this code working, which converts from hex to base64, and vice versa. I got to_base64 from another SO question, and I wrote to_hex with some guesswork and trial and error.

class String

  def to_base64
    [[self].pack("H*")].pack("m0")
  end

  def to_hex
    self.unpack("m0").first.unpack("H*").first
  end
end

但是，即使在阅读了文档之后，我也不真正使用pack和unpack方法.具体来说，我对两个实现之间的不对称感到困惑.从概念上讲，在两种情况下，我们都采用以某个基数(16或64)编码的字符串，并希望将其转换为另一个基数.那么为什么我们不能像这样实现to_hex:

But I don't really grok the pack and unpack methods, even after reading the docs. Specifically, I'm confused by the asymmetry between the two implementations. Conceptually, in both cases, we take a string encoded in some base (16 or 64), and we wish to convert it to another base. So why can't we implement to_hex like this:

def to_hex
  [[self].pack("m0")].pack("H*")
end

或to_base64使用unpack?为什么我们选择的基础完全改变了完成转换所需的方法?

or to_base64 using unpack? Why does the base we chose completely change the method we need to use to accomplish conversions?

推荐答案

to_hex是to_base64的确切逆:

1. 将字符串放入数组:[self]
2. 带有H*的呼叫包:[self].pack("H*")
3. 将字符串放入数组中:[[self].pack("H*")]
4. 带有m0的呼叫包:[[self].pack("H*")].pack("m0")

1. put string in an array: [self]
2. call pack with H*: [self].pack("H*")
3. put string in an array: [[self].pack("H*")]
4. call pack with m0: [[self].pack("H*")].pack("m0")

to_hex

1. 使用m0呼叫解压缩:self.unpack("m0")
2. 从数组中提取字符串:self.unpack("m0").first
3. 使用H*调用拆包:self.unpack("m0").first.unpack("H*")
4. 从数组中提取字符串:self.unpack("m0").first.unpack("H*").first

1. call unpack with m0: self.unpack("m0")
2. extract string from array: self.unpack("m0").first
3. call unpack with H*: self.unpack("m0").first.unpack("H*")
4. extract string from array: self.unpack("m0").first.unpack("H*").first

这就是通过应用反操作来撤消操作的方式:

That's how you undo operations, by applying the inverse operations:

a = 5
(a + 4) * 3
#=> 27

反之亦然:

a = 27
(a / 3) - 4
#=> 5

a.pack是a.unpack的逆，而a.first是[a]

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