为什么是“$ _”在一个Perl一行程序一样$ ARGV？ [英] Why is '$_' the same as $ARGV in a Perl one-liner?

问题描述

我就遇到了这个问题，同时试图一个Perl的一个班轮打印单引号。我终于想通了，你有来逃避他们'\\''。下面是一些code来说明我的问题。

I ran into this problem while trying to print single quotes in a Perl one-liner. I eventually figured out you have to escape them with '\''. Here's some code to illustrate my question.

让我们先从打印的文本文件。

Let's start with printing a text file.

perl -ne 'chomp; print "$_\n"' shortlist.txt

red
orange
yellow
green
blue

现在让我们来打印文件的名称，而不是为每一行。

Now let's print the name of the file instead for each line.

perl -ne 'chomp; print "$ARGV\n"' shortlist.txt

shortlist.txt
shortlist.txt
shortlist.txt
shortlist.txt
shortlist.txt

然后，我们可以在每行加单引号。

Then we can add single quotes around each line.

perl -ne 'chomp; print "'$_'\n"' shortlist.txt

shortlist.txt
shortlist.txt
shortlist.txt
shortlist.txt
shortlist.txt

等待，没有工作。让我们再试一次。

Wait that didn't work. Let's try again.

perl -ne 'chomp; print "'\''$_'\''\n"' shortlist.txt

'red'
'orange'
'yellow'
'green'
'blue'

所以我现在的工作。但我为什么$ _计算程序名称仍然感到困惑。也许这是一件很容易，但有人可以解释或链接到一些文档？

So I got it working now. But I'm still confused on why '$_' evaluates to the program name. Maybe this is something easy but can someone explain or link to some documentation?

编辑：我在Red Hat 5上运行的Perl 5.8.8

edit: I'm running Perl 5.8.8 on Red Hat 5

推荐答案

要你的shell，格格;打印'$ _的'\\ n'''/ code>后的字符串这是并置

To your shell, 'chomp; print "'$_'\n"' results in a string that's the concatenation of

1. 格格;打印（单引号里面的第一个序列），


2. 的变量 $ _ 的值，


3. \\ n（单引号内的第二序列）。

1. chomp; print " (the first sequence inside single quotes),
2. the value of its variable $_, and
3. \n" (the second sequence inside single quotes).

在庆典， $ _ ...扩展到最后一个参数previous命令膨胀后......。由于这恰好是 shortlist.txt ，被传递到 perl的以下内容：

In bash, $_ "... expands to the last argument to the previous command, after expansion. ...". Since this happens to be shortlist.txt, the following is passed to perl:

chomp; print "shortlist.txt\n"

例如，

$ echo foo
foo

$ echo 'chomp; print "'$_'\n"'
chomp; print "foo\n"

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