在Clojure中修改Clojure源代码文件 [英] Modify Clojure source code file in clojure

问题描述

我想知道是否可以将Clojure .clj 源文件中包含的代码作为列表加载而不进行编译。

I was wondering if it's possible to load the code contained in a Clojure .clj source file as a list, without compiling it.

如果我可以将 .clj 文件作为列表加载，则可以修改该列表并将其漂亮地打印回同一文件中，然后

If I can load a .clj file as a list, I can modify that list and pretty print it back into the same file which can then be loaded again.

（也许这是一个坏主意。）有人知道这是否可能吗？

(Maybe this is a bad idea.) Does anyone know if this is possible?

推荐答案

一个简单的示例：

user=> (def a '(println (+ 1 1))) ; "'" escapes the form to prevent immediate evaluation
#'user/a
user=> (spit "test.code" a) ; write it to a file
nil

user=> (def from-file (read-string (slurp "test.code"))) ; read it from a file
#'user/from-file
user=> (def modified (clojure.walk/postwalk-replace {1 2} from-file)) ; modify the code
#'user/modified
user=> (spit "new.code" modified) ; write it back
nil
user=> (load-string (slurp "new.code")) ; check it worked!
4
nil

在哪里 lur口给您一个字符串， read-string 给您一个未求值的表格，而 load-string 给您评估表单的结果。

Where slurp gives you a string, read-string gives you an un-evaluated form, and load-string gives you the result of evaluating the form.

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