问题描述
我最近一直在玩clojure，并遇到一个问题，我不知道如何处理。我有一个doseq有7个参数，它扩展到一个巨大的块，几乎通过最大类的大小。为什么doseq扩展到这么大的clojure代码块？
I've recently been playing with clojure and reached a problem that I'm not sure how to handle. I have a doseq with 7 parameters and it expands to a huge block, almost passing the maximum class size. Why does doseq expand to such a huge block of clojure code?
示例：
(def q '(doseq
[p0 (nth (:params operator) 0 (quote (nil)))
p1 (nth (:params operator) 1 (quote (nil)))
p2 (nth (:params operator) 2 (quote (nil)))
p3 (nth (:params operator) 3 (quote (nil)))
p4 (nth (:params operator) 4 (quote (nil)))
p5 (nth (:params operator) 5 (quote (nil)))
p6 (nth (:params operator) 6 (quote (nil)))]
(do-print board (:oname operator) p0 p1 p2 p3 p4 p5 p6)))
然后：
(macroexpand q)
在我的机器上，这给出了一大块代码（97331字节）。这是正常还是我做错了什么？运算符是一个简单的defrecord。如果任何人感兴趣，可以点击以下链接查看扩展结果：
on my machine this gives a huge chunk of code (97331 bytes). Is this normal or am I doing something wrong? Operator is a simple defrecord. Here's a link to the expanded result if anyone is interested: http://pastebin.com/6gw1q078
编辑：
通过执行相同操作，但使用 for 数量级小（3653字节）：
By doing the same, but with a for form I get something a few orders of magnitude smaller (3653 bytes):
(def q '(for
[p0 (nth (:params operator) 0 (quote (nil)))
p1 (nth (:params operator) 1 (quote (nil)))
p2 (nth (:params operator) 2 (quote (nil)))
p3 (nth (:params operator) 3 (quote (nil)))
p4 (nth (:params operator) 4 (quote (nil)))
p5 (nth (:params operator) 5 (quote (nil)))
p6 (nth (:params operator) 6 (quote (nil)))]
(do-print board (:oname operator) p0 p1 p2 p3 p4 p5 p6)))
(macroexpand q)
：
为什么两者之间有这么大的差别？ doseq形式看起来没有问题，当我收到一个错误，说超过了java类大小时，我真的很惊讶。
Why is there such a huge difference between the two? The doseq form looks innocent enough, I was really surprised when I got an error saying that java class size has been exceeded.
推荐答案
好了，看一下较小的 doseq 宏观消费的原因：
Well, looking at a smaller macro-expasion of doseq reveals the reason:
(loop [seq_2365 (seq [1 2])
chunk_2366 nil
count_2367 0
i_2368 0]
(if (< i_2368 count_2367)
(let [x (.nth chunk_2366 i_2368)]
(do x)
(recur seq_2365 chunk_2366 count_2367 (unchecked-inc i_2368)))
(when-let [seq_2365 (seq seq_2365)]
(if (chunked-seq? seq_2365)
(let [c__4197__auto__ (chunk-first seq_2365)]
(recur (chunk-rest seq_2365) c__4197__auto__ (int (count c__4197__auto__)) (int 0)))
(let [x (first seq_2365)]
(do x)
(recur (next seq_2365) nil 0 0))))))
理想情况下，我们只需要最后的 let doseq 正在发出额外的代码，以特定的方式处理 chunked-seq ，使得它占用第一个块，然后块中的每个项目执行 doseg 的正文。
Ideally we only need the last let form but doseq is emitting additional code to handle chunked-seq in a specific way such that it takes the first chunk and then for each item in the chunk execute the body of doseq and so on.
此代码是为 doseq 但是当你有第二个seq，那么类似的代码处理chunked-seq产生，因此它的大小爆炸。
This code is generated for a single seq in doseq but when you have 2nd seq in it then similar code for handling chunked-seq is generated and hence it explod in size.
这篇关于Clojure doseq生成巨大的代码？的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持！