Clojure关键字的内存大小是多少？

Question

I'm writing a function in Clojure to estimate the in-memory size of a parsed JSON, something like:

(defn object-size
  [object]
  (cond
    (sequential? object)
      (reduce + (map object-size object))
    (map? object)
      (reduce
        (fn [total [k v]]
          (+ total (keyword-size k) (object-size v)))
        0
        object)
    :else
      (case (type object)
        java.lang.Long 8 
        java.lang.Double 8
        java.lang.String (* 2 (count object))
        ;; other data types
      )))

Obviously I'll need to add in overheads for clojure.lang.PersistentVector , java.lang.String , etc.

However, I'm not sure how to find the in-memory size of a clojure.lang.Keyword , the keyword-size function in the above example. How does Clojure store keywords? Are they constant size similar to a C++ enum , or are they a special case of java.lang.String that are dependent on length?

Answer 1

Answering this question from within Clojure is basically impossible. Your first-draft function works okay for the very simplest data structures, although even this simplest attempt has several errors already.

But more than that, it is just an ill-framed question. What is the size of xs in this snippet?

(def xs (let [forever (promise)]
          (deliver forever
                   (lazy-seq (cons 1 @forever)))
          @forever))

user=> (take 5 xs)
(1 1 1 1 1)

xs is an infinitely long sequence (so your reduce will never complete, but if it could it would surely return "this is infinite"). But it actually takes a small, fixed amount of memory, because it is circular.

You may say, well gee this is a dumb object, I don't mind if my function fails for objects like that. But in a garbage-collected language with pervasive laziness, cases with similar characteristics are commonplace. If you rule them out, you rule out everything interesting.