问题描述
cat 不从管道输入中获取文件名列表的设计原理是什么?为什么设计师选择以下方法不起作用?
What is the design rationale that cat doesn't take list of file names from pipe input? Why did the designers choose that the following does not work?
ls *.txt | cat
相反，他们选择我们需要将文件名作为参数传递给 cat 为:
Instead of this, they chose that we need to pass the file names as argument to cat as:
ls *.txt | xargs cat
推荐答案
当你说 ls *.txt |cat 不起作用，你应该说它不像你期望的那样工作.事实上，它的工作方式被认为是有效的.
When you say ls *.txt | cat doesn't work, you should say that doesn't work as you expect. In fact, that works in the way it was thought to work.
来自man:
cat - Concatenate FILE(s), or standard input, to standard output
假设下一个输出:
$ ls *.txt
file1.txt
file2.txt
... cat 的输入将是:
... the input to cat will be:
file1.txt
file2.txt
...这正是cat在标准输出
在某些 shell 中，它相当于:
In some shells, it's equivalent to:
cat <(ls *.txt)
或
ls *.txt > tmpfile; cat tmpfile
所以，cat 真的像他们的设计师所期望的那样工作.
So, cat is really working as their designers expected to do so.
另一方面，您所期望的是 cat 将其输入解释为一组文件名以读取和连接它们的内容，但是当您 pipe 到 cat，该输入作为一个单独的文件.
On the other hand, what you are expecting is that cat interprets its input as a set of filenames to read and concatenate their content, but when you pipe to cat, that input works as a lonely file.
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