Elixir：将位列表转换为二进制 [英] Elixir: Convert list of bits to binary

问题描述

我有一个代表位的整数列表；例如 [1,0,0,1,1,0,0,1,0,0,0,1,1,1,0,0] ，我想要将其转换为二进制文件，即< 153，28> ，我知道列表的长度将始终是8的倍数。

I have a list of integers representing bits; e.g. [1,0,0,1,1,0,0,1,0,0,0,1,1,1,0,0] and I would like to convert it into a binary i.e. <<153, 28>>, I know that the length of the list will always be a multiple of 8.

我查看了Elixir文档，但未能找到任何帮助（我查找了确切的函数，但也查找了向二进制文件追加一点的函数）。

I have looked at the Elixir documentation, but I have not been able to find any help (I have looked for the exact function but also for a function for appending a bit to a binary).

我写了一个可以解决问题的函数（如下），但我希望有一种更好的方法，因为我认为我的函数看起来太复杂了。

I have written a function which solves the problem (below) but I hoped there was a better way as I think my function looks too complicated.

def list_to_binary(l) do 
  if length(l) >= 8 do
    << Enum.at(l, 0) :: size(1),
      Enum.at(l, 1) :: size(1),
      Enum.at(l, 2) :: size(1),
      Enum.at(l, 3) :: size(1),
      Enum.at(l, 4) :: size(1),
      Enum.at(l, 5) :: size(1),
      Enum.at(l, 6) :: size(1),
      Enum.at(l, 7) :: size(1)
    >> <> list_to_binary(Enum.drop l, 8)
  else
    if length(l) == 0 do
      <<>>
    else
      l = l ++ List.duplicate(0, 8 - length(l))
      list_to_binary(l)
    end
  end
end

推荐答案

使用 Kernel.SpecialForms.for / 1 理解：它是 into 关键字参数接受实现 可收集 协议的任何内容和 binary 确实实现了它。

Use Kernel.SpecialForms.for/1 comprehension: it’s into keyword argument accepts anything implementing Collectable protocol and binary indeed does implement it.

for i <- [1,0,0,1,1,0,0,1,0,0,0,1,1,1,0,0], do: <<i::1>>, into: <<>>
#⇒ <<153, 28>>

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