“&”如何操作员工作？ [英] How does the '&' operator work?

问题描述

在以下情况下，我无法理解'&'运算符。

I am having trouble understand the '&' operator in the following context.

  @doc "Marks a task as executed"
  def put_task(task, project) do
    item = {task, project}
    Agent.update(__MODULE__, &MapSet.put(&1, item))
  end

在这种情况下，& 1似乎是指地图对象本身，但是我对此工作方式感到好奇。我是否在文档中对此进行了讨论，但无法确定这是否正是事实呢？如果有人可以帮助我了解到底发生了什么以及& 1所指的内容，以及如果它所指的是MapSet的话，这是怎么可能的，我将不胜感激。

It seems that in this case the '&1' is referring to the map object itself, but I am curious as to how this works. Is it passing itself in as an argument I looked into this in the docs but couldn't find out if this was exactly what was going on. I would be grateful if somebody could help me understand what is exactly going on and what &1 refers to and if it refers to the MapSet how is this possible.

推荐答案

如果有人可以帮助我了解发生了什么，以及; 1引用，如果引用了MapSet，这怎么可能。

I would be grateful if somebody could help me understand what is exactly going on and what &1 refers to and if it refers to the MapSet how is this possible.

Agent.update / 3 调用给定函数当前状态，并将返回值存储为新状态。由于& MapSet.put（& 1，item）与 fn x->相同， MapSet.put（x，item）结尾， x 在这里变为旧状态，而新的 MapSet MapSet.put / 2 返回的c $ c>成为代理的新状态。

Agent.update/3 calls the given function with the current state, and stores the returned value as the new state. Since &MapSet.put(&1, item) is the same as fn x -> MapSet.put(x, item) end, x here becomes the old state, and the new MapSet returned by MapSet.put/2 becomes the new state of the agent.

对于此代码要正常运行，必须使用名称为的 Agent.start 或 Agent.start_link 进行调用：__MODULE __ 和返回 {：ok，map_set} 的函数，其中 map_set 是任意值 MapSet 在代码中的某个位置。

For this code to function, there must be a call to Agent.start or Agent.start_link with name: __MODULE__ and a function that returns {:ok, map_set} where map_set is any MapSet somewhere in the code.

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