将列表上下文结果转换为 perl 中一行的数组? [英] Conversion of list context result to array in one line in perl?

问题描述

我用 perl 写了这段代码:

I wrote this code in perl:

shift( @interfaces = qx'ifconfig -s' );

并得到这个错误:

Type of arg 1 to shift must be array (not list assignment)

当我这样写时:

@interfaces = qx'ifconfig -s';
shift @interfaces;

它完成了我想要的，即以行数组的形式获取 ifconfig 命令的输出并删除数组中的第一个元素(这是一个标题，而不是实际的接口).

It does what I want, which is to get the output of the ifconfig command as an array of lines and remove the first element in the array (which is a header, not an actual interface).

我个人的偏好是将其写成单行.在我看来，第一个例子中的括号应该导致赋值被完全解析，因此允许 shift 将 @interfaces 视为一个数组，但很明显 perl 认为它是一个列表赋值".

My personal preference is to write this as a one liner. It seems to me the parentheses in the first example should cause the assignment to be resolved fully, therefore allowing shift to see @interfaces as an array, but clearly perl thinks it's a "list assignment."

这对于 perl 大师来说肯定是一个简单的问题，但我已经用谷歌搜索了又谷歌搜索并没有找到启示.

This is surely an easy question for the perl gurus but I've googled and googled and haven't found enlightenment.

如果有人愿意提供特定的语义以在一行中完成我想要的内容，我将不胜感激.如果您也愿意花时间解释为什么我的第一个版本不起作用，我将永远感激(教一个人钓鱼等等).

If someone would please provide the specific semantics to accomplish what I want in one line I would appreciate it. If you would also please take the time to explain why my first version isn't working I would be eternally grateful (teach a man to fish and all that).

预先感谢您的帮助.

推荐答案

如您所见，shift 需要一个文字数组，而不是赋值的结果.这是因为当 perl 解析 shift @interfaces 时，它实际上是将它改写成类似 &CORE::shift(\@interfaces) 的东西，你不能参考一个赋值并获得一个数组引用.

As you saw, shift requires a literal array, not the result of an assignment. This is because when perl parses shift @interfaces it is actually rewriting it into something like &CORE::shift(\@interfaces) and you can not take a reference of an assignment and get an array ref.

您可以将它分成两行，如mob所示，您可以将赋值隐藏在括号内的取消引用中，或者您可以简单地丢弃第一个值:

You could break it into two lines as you have found, you could hide the assignment inside a bracketed dereference as mob shows, or you could simply throw away the first value:

(undef, @interfaces) = qx'ifconfig -s';

undef 在左值位置是不需要的值的占位符.

undef in an lvalue position is a placeholder for values that you don't need.

(shift 的解析在 perl 5.14+ 中发生了一些变化，但上面的参数仍然成立)

(parsing of shift has changed a bit in perl 5.14+, but the argument above still holds)

还有一些你可能不应该使用的方法，只能通过增加长度来排序:)

a few more ways that you probably shouldn't use, ordered only by increasing length :)

my @interfaces = sub {shift; @_}->(qx'ifconfig -s');

my @interfaces = sub {@_[1..$#_]}->(qx'ifconfig -s');

my @interfaces = map {@$_[1..$#$_]} [qx'ifconfig -s'];

my @interfaces = map {shift @$_; @$_} [qx'ifconfig -s'];

our @interfaces; shift @{*interfaces = [qx'ifconfig -s']};

my @interfaces = sub {*_ = [qx'ifconfig -s']; shift; @_}->();

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