使用`line-seq`和`reader`，什么时候文件关闭？ [英] Using `line-seq` with `reader`, when is the file closed?

问题描述

我正在使用（line-seq（readerinput.txt））从文本文件中读取行。

I'm reading lines from a text file using (line-seq (reader "input.txt")). This collection is then passed around and used by my program.

我担心这可能是不好的风格，因为我不是确定性地关闭文件。我想我不能使用（with-open（line-seq（readerinput.txt））），因为文件流可能会在我之前关闭

I'm concerned that this may be bad style however, as I'm not deterministically closing the file. I imagine that I can't use (with-open (line-seq (reader "input.txt"))), as the file stream will potentially get closed before I've traversed the entire sequence.

应避免与结合使用 lazy-seq 阅读器文件？是否有不同的模式我应该在这里？

Should lazy-seq be avoided in conjunction with reader for files? Is there a different pattern I should be using here?

推荐答案

因为这不是真的有一个明确的答案第一个答案的注释），这里是它的本质：

Since this doesn't really have a clear answer (it's all mixed into comments on the first answer), here's the essence of it:

(with-open [r (reader "input.txt")]
  (doall (line-seq r)))

要读取和关闭文件的整行序列。

That will force the whole sequence of lines to be read and close the file. You can then pass the result of that whole expression around.

当处理大型文件时，你可能会遇到内存问题（在内存中保存整行序列），当转换程序是一个好主意：

When dealing with large files, you may have memory problems (holding the whole sequence of lines in memory) and that's when it's a good idea to invert the program:

(with-open [r (reader "input.txt")]
  (doall (my-program (line-seq r))))

在这种情况下可能不需要doall，这取决于我的程序返回和/或我的程序是否懒惰地消耗序列。

You may or may not need doall in that case, depending on what my-program returns and/or whether my-program consumes the sequence lazily or not.

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