Clojure的defrecord - 如何使用它？ [英] Clojure's defrecord - how to use it?

问题描述

我试图在Clojure中使用 defrecord 创建我自己的不可变数据类型/方法。目标是有一个数据类型，我可以创建的实例，然后调用其方法来返回一个带有变量变量的新副本。说a和b是向量。我想更新一个值，并返回一个新的副本的整个结构与这些向量更新。这显然不编译，我只是想得到我的想法。

I'm attempting to create my own immutable datatype/methods with defrecord in Clojure. The goal is to have a datatype that I can create instances of, and then call its methods to return a new copy of itself with mutated variables. Say a and b are vectors. I'd like to update a value in both and return a new copy of the entire structure with those vectors updated. This obviously doesn't compile, I'm just trying to get my ideas across.

(defrecord MyType [a b]
  (constructor [N]
    ; I'd like to build an initial instance, creating a and b as vectors of length N
  ) 

  (mutate-and-return [] 
    ; I'd like to mutate (assoc the vectors) and return the new structure, a and b modified
  )
)

我想调用构造函数，然后调用mutator多次，我想要的（还有其他函数不变异，但我

I'd like to call the constructor and then the mutator as many times as I'd like (there are other functions that don't mutate, but I don't want to make it more complex for the question).

或者，如果这不是惯用的Clojure，你应该怎么做这样的事情？

Alternatively, if this is not idiomatic Clojure, how are you supposed to do something like this?

推荐答案

以下是您如何定义记录：

Here's how you define your record:

(defrecord MyType [a b])

请注意，在Clojure中，

Note that in Clojure you don't typically define "methods" within your record type itself (the exception is if you want to directly implement a Java interface or a protocol).

一个基本的构造函数（以<$ c为前缀） $ c> - > ）免费自动生成：

A basic constructor (prefixed with ->) gets generated automatically for free:

(def foo (->MyType [1 2 3] [4 5 6]))

foo
=> #user.MyType{:a [1 2 3], :b [4 5 6]}

然后你可以写更复杂的构造函数，使用这个，例如

You can then write more sophisticated constructor functions that use this, e.g.

(defn mytype-with-length [n]
  (let [a (vec (range n))
        b (vec (range n))] 
    (->MyType a b)))

(mytype-with-length 3)
=> #user.MyType{:a [0 1 2], :b [0 1 2]}

和mutate-and-return也是免费的 - 您只需使用 assoc ：

And "mutate-and-return" also comes for free - you can just use assoc:

(assoc foo :b [7 8 9])
=> user.MyType{:a [1 2 3], :b [7 8 9]}

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