将元数据添加到延迟序列

Question

当我尝试在Clojure中向无限延迟序列添加元数据时,我得到了堆栈溢出,如果我取消元数据,那么它可以正常工作.为什么添加with-meta宏打破了懒惰的seq？

首先创建一个非常好的数字的无限序列:

(defn good []
  (lazy-seq 
    (cons 42
      (good))))

user> (take 5 (good))
(42 42 42 42 42)

然后,为每个lazy-seq实例添加一些元数据:

(defn bad []
  (lazy-seq 
    (cons 42
      (with-meta 
       (bad)
       {:padding 4}))))


user> (take 5 (bad))
java.lang.StackOverflowError (NO_SOURCE_FILE:0)
  [Thrown class clojure.lang.Compiler$CompilerException]

尝试将元数据上移一级:

(defn also-bad []
  (with-meta 
   (lazy-seq 
     (cons 42
       (also-bad)))
   {:padding 4}))

user> (take 5 (foo))
java.lang.StackOverflowError (NO_SOURCE_FILE:0)
  [Thrown class clojure.lang.Compiler$CompilerException]

以下是有限序列的元数据示例:

(defn also-works []
     (lazy-seq 
       (cons 4 
         (with-meta 
          () 
          {:a 5}))))

user> (also-works)
(4)
user> (meta (rest (also-works)))
{:a 5}
user> 

Answer 1

因为LazySeq只要你拨打评估其身体withMeta上LazySeq.你失去了懒惰.

public final class LazySeq extends Obj implements ISeq, List{
    ...
    public Obj withMeta(IPersistentMap meta){
        return new LazySeq(meta, seq());
    }
    ...
}

seq()如果尚未评估懒惰seq的主体,则对其进行评估.上面的代码不断调用with-meta连续的lazy seqs,它们会一直评估它们,直到堆栈爆炸.我认为目前没有任何方法可以将元数据添加到惰性seq而不会导致它评估其正文.