为什么在管道中读取和写入同一文件会产生不可靠的结果? [英] Why does reading and writing to the same file in a pipeline produce unreliable results?

问题描述

我有一堆文件，其中包含许多空白行，并且想要删除任何重复的空白行以使读取文件更加容易.我写了以下脚本:

I have a bunch a files that contain many blank lines, and want to remove any repeated blank lines to make reading the files easier. I wrote the following script:

#!/bin/bash
for file in * ; do cat "$file" | sed 's/^ \+//' | cat -s > "$file" ; done

但是，这产生了非常不可靠的结果，大多数文件完全变为空，只有少数文件具有预期的结果.更重要的是，每次重试时，起作用的文件似乎都是随机变化的，因为每次运行都会对不同的文件进行正确的编辑.发生了什么事?

However, this had very unreliable results, with most files becoming completely empty and only a few files having the intended results. What's more, the files that did work seemed to change randomly every time I retried, as different files would get correctly edited in every run. What's going on?

注意:这更多是理论上的问题，因为我意识到我可以使用类似的解决方法:

Note: This is more of a theoretical question, because I realize I could use a workaround like:

#!/bin/bash
for file in * ; do 
    cat "$file" | sed 's/^ \+//' | cat -s > "$file"-tmp
    rm "$file"
    mv "$file"-tmp "$file"
done

但这似乎不必要地令人费解.那么，为什么直接"方法如此不可靠?

But that seems unnecessarily convoluted. So why is the "direct" method so unreliable?

推荐答案

发生不可预测性是因为在管道中的两个阶段cat "$file"和cat -s > "$file"之间存在竞争条件.

The unpredictability happens because there's a race condition between two stages in the pipeline, cat "$file" and cat -s > "$file".

第一个尝试打开文件并从中读取，而另一个尝试清空文件.

The first tries to open the file and read from it, while the other tries to empty the file.

• 如果在读取之前将其清空，则会得到一个空文件.
• 如果在清空之前将其读取，您会得到一些数据(但是文件很快就会清空，除非结果很短，否则结果将被截断).

如果您具有GNU sed，则只需执行sed -i 'expression' *

If you have GNU sed, you can simply do sed -i 'expression' *

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