Clojure：简单的阶乘导致堆栈溢出 [英] Clojure: Simple factorial causes stack overflow

问题描述

我做错了什么？简单递归几千个调用深度抛出一个 StackOverflowError 。

What am I doing wrong? Simple recursion a few thousand calls deep throws a StackOverflowError.

如果Clojure递归的限制如此低，我可以依靠它吗？

If the limit of Clojure recursions is so low, how can I rely on it?

(defn fact[x]
  (if (<= x 1) 1 (* x  (fact (- x 1))  )))

user=> (fact 2)
2

user=> (fact 4)
24

user=> (fact 4000)
java.lang.StackOverflowError (NO_SOURCE_FILE:0)

推荐答案

堆栈大小，我明白，取决于你使用的JVM以及平台。如果使用Sun JVM，可以使用 -Xss 和 -XThreadStackSize 参数来设置堆栈大小。

The stack size, I understand, depends on the JVM you are using as well as the platform. If you are using the Sun JVM, you can use the -Xss and -XThreadStackSize parameters to set the stack size.

在Clojure中，递归的首选方法是使用 loop / recur ：

The preferred way to do recursion in Clojure though is to use loop/recur:

(defn fact [x]
    (loop [n x f 1]
        (if (= n 1)
            f
            (recur (dec n) (* f n)))))

Clojure将为此进行尾调用优化;确保你不会遇到 StackOverflowError 。

Clojure will do tail-call optimization for this; that ensures that you’ll never run into StackOverflowErrors.

编辑：为了记录，这里是一个Clojure函数返回所有阶乘的延迟序列：

For the record, here is a Clojure function that returns a lazy sequence of all the factorials:

(defn factorials []
    (letfn [(factorial-seq [n fact]
                           (lazy-seq
                             (cons fact (factorial-seq (inc n) (* (inc n) fact)))))]
      (factorial-seq 1 1)))

(take 5 (factorials)) ; will return (1 2 6 24 120)

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