How does clojure bind variable parameters?

Question

I am a newbie in Clojure.The problem originated when I once checked the source code of conj:

    (def conj 
      (fn ^:static conj
        ([] [])
        ([coll] coll)
        ([coll x] (clojure.lang.RT/conj coll x));4
        ([coll x & xs] ;1
         (if xs ;2
           (recur (clojure.lang.RT/conj coll x) (first xs) (next xs)) ;3
           (clojure.lang.RT/conj coll x)))))

The source code of conj shows that it uses recur to implement the function. This source code looks very simple. What I'm confused about is the condition it uses when determining whether recursion needs to continue. It looks like it checks if the variable parameters is nil , but if the variable parameters is nil, it will soon be equivalent to the third "arity" of conj? I then tried to evaluate the following expression:

user=> (conj [] 1 (next []))
[1 nil]
user=>

It works normally and successfully adds nil to the vector. I understand that clojure actually wraps nil in a list and passes it to the function, but I don't understand why recur can pass a real nil in ? And why does clojure recognize and match the correct "arity"?

user=> (def my_conj
   (fn [coll x & xs]
     (println "xs is" xs)
     (if xs
       (recur (clojure.lang.RT/conj coll x) (first xs) (next xs))
       (clojure.lang.RT/conj coll x))))
#'user/my_conj
user=> (my_conj [] 1 (next []))
xs is (nil)
xs is nil
[1 nil]

andy_fingerhut · Accepted Answer · 2019-12-24 22:09:20Z

1

OK, my apologies for not realizing earlier that you have come across an aspect of Clojure behavior that I had never seen before. Today I learned something new about Clojure, and given 10 years with it, that surprised me.

This is mentioned in a couple of sentences of the official Clojure documentation for recur, here: https://clojure.org/reference/special_forms#recur

Here is a page of community-written examples and documentation (so not "official") that describes this behavior of recur with functions that have variadic arguments: https://clojuredocs.org/clojure.core/recur#example-55ff3cd4e4b08e404b6c1c7f

answered Dec 24, 2019 at 22:09

andy_fingerhut

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andy_fingerhut · Accepted Answer · 2019-12-24 04:29:40Z

0

I would recommend making a copy of this function, changing its name to something else, and adding some calls to println of the values of whatever interests you, e.g. x, xs, etc, and watch what gets printed in a Clojure REPL session when you call your function with different parameters that interest you.

The if xs is true if the value of xs is any value at all other than nil or false. The value (nil) is a list of one element, and is considered true when used as an if conditional expression.

answered Dec 24, 2019 at 4:29

andy_fingerhut

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6 Comments

两极化 Over a year ago

Thanks for your answer, but I actually tried to do this before posting this question, and the result is the same as described in the question.

andy_fingerhut Over a year ago

(next []) evaluates to nil, so your example call (my_conj [] 1 (next []))) is the same as if you had called (my_conj [] 1 nil).

andy_fingerhut Over a year ago

Also I do not know if this is confusing, but clojure.lang.RT/conj is a call to a Java method, not a recursive call.

两极化 Over a year ago

(recur (clojure.lang.RT/conj coll x) (first xs) (next xs)) ;3 Isn't this a recursive call?And it should be simplified to:(recur result (first xs) nil),result is the evaluation result of (clojure.lang.RT/conj coll x), nil is (next xs) because xs contains only one element: nil.

andy_fingerhut Over a year ago

One more thing, which I am not 100% certain about, but I believe is true: When you call conj/my-conj with exactly 2 arguments, it looks as if Clojure could call either the version declared with parameters [coll x] or [coll x & xs], but it appears to always call the one declared with [coll x] in that case, when called from outside. However, the recur call will never switch between different declarations -- it will always call the definition with the same arity, no matter what. I believe that is a limitation/property of recur in Clojure.

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andy_fingerhut · Accepted Answer · 2019-12-24 07:27:53Z

0

The part of the function declared with parameters [coll x & xs] means "bind the value of the first parameter to coll, bind the value of the second parameter to x, then if there are no more parameters, bind xs with the value nil, otherwise bind xs with the value that is the list of the remaining parameters".

You can see this with this simpler function that does not use recur at all:

user=> (defn my-fn [a & xs]
         (println "a=" a " xs=" xs))

user=> (my-fn 1)
a= 1  xs= nil

user=> (my-fn 1 2)
a= 1  xs= (2)

answered Dec 24, 2019 at 7:27

andy_fingerhut

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1 Comment

两极化 Over a year ago

But how to evaluate this form: (recur (clojure.lang.RT/conj coll x) (first xs) (next xs)) ;3?Unlike your example, recur really declares additional parameters! If you use this example to explain it, it may end up being evaluated as (recur (clojure.lang.RT/conj coll x) (first xs)) ;3. Because the last parameter is "nil", it will not be passed. I think it makes sense to explain this, maybe recur did something,but I can't find other information to prove it..

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