'concat'打破了'line-seq'的懒惰吗？ [英] Does 'concat' break the laziness of 'line-seq'?

问题描述

以下代码似乎强制 line-seq 从文件中读取4行。这是一种缓冲机制吗？我需要在这里使用 lazy-cat 吗？如果是这样，我如何将宏应用到序列，就像它是可变参数一样？

The following code appears to force line-seq to read 4 lines from file. Is this some kind of buffering mechanism? Do I need to use lazy-cat here? If so, how can I apply a macro to a sequence as if it were variadic arguments?

(defn char-seq [rdr]
  (let [coll (line-seq rdr)]
    (apply concat (map (fn [x] (println \,) x) coll))))

(def tmp (char-seq (clojure.contrib.io/reader file)))

;,
;,
;,
;,
#'user/tmp

推荐答案

你看到的部分是由于 apply ，因为它需要实现函数定义所需的多个参数。例如：

Part of what you're seeing is due to apply, since it will need to realize as many args as needed by the function definition. E.g.:

user=> (defn foo [& args] nil)
#'user/foo
user=> (def bar (apply foo (iterate #(let [i (inc %)] (println i) i) 0)))
1
#'user/bar
user=> (defn foo [x & args] nil)
#'user/foo
user=> (def bar (apply foo (iterate #(let [i (inc %)] (println i) i) 0)))
1
2
#'user/bar
user=> (defn foo [x y & args] nil)
#'user/foo
user=> (def bar (apply foo (iterate #(let [i (inc %)] (println i) i) 0)))
1
2
3
#'user/bar

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