如何在 perl 中创建多维数组? [英] How can I create a multi-dimensional array in perl?

问题描述

我以这种方式创建了一个多维数组:

I was creating a multi-dimensional array this way:

#!/usr/bin/perl
use warnings;
use strict;

my @a1 = (1, 2);
my @a2 = (@a1, 3);

但结果我还是得到了一个一维数组...

But it turns out that I still got a one-dimensional array...

Perl 的正确方法是什么?

What's the right way in Perl?

推荐答案

你得到一个一维数组，因为数组 @a1 在括号内展开.所以，假设:

You get a one-dimensional array because the array @a1 is expanded inside the parentheses. So, assuming:

my @a1 = (1, 2);
my @a2 = (@a1, 3);

那么你的第二个语句等价于my @a2 = (1,2,3);.

Then your second statement is equivalent to my @a2 = (1,2,3);.

创建多维数组时，您有几个选择:

When creating a multi-dimensional array, you have a few choices:

1. 直接分配每个值
2. 取消引用现有数组
3. 插入引用

第一个选项基本上是$array[0][0] = 1;，不是很精彩.

The first option is basically $array[0][0] = 1; and is not very exciting.

第二个是这样做的:my @a2 = (\@a1, 3);.请注意，这会引用数组 @a1 的命名空间，因此如果您稍后更改 @a1，@a2 中的值将也改变.它并不总是推荐的选项.

The second is doing this: my @a2 = (\@a1, 3);. Note that this makes a reference to the namespace for the array @a1, so if you later change @a1, the values inside @a2 will also change. It is not always a recommended option.

第二个选项的变体是这样做:my @a2 = ([1,2], 3);.括号将创建一个匿名数组，该数组没有命名空间，只有内存地址，并且只存在于@a2 中.

A variation of the second option is doing this: my @a2 = ([1,2], 3);. The brackets will create an anonymous array, which has no namespace, only a memory address, and will only exist inside @a2.

第三个选项，有点模糊，是这样做的: my $a1 = [1,2];我的@a2 = ($a1, 3);.它将做与 2 完全相同的事情，只是数组引用已经在一个名为 $a1 的标量变量中.

The third option, a bit more obscure, is doing this: my $a1 = [1,2]; my @a2 = ($a1, 3);. It will do exactly the same thing as 2, only the array reference is already in a scalar variable, called $a1.

注意() 和[] 在赋值给数组时的区别.括号 [] 创建一个匿名数组，它返回一个数组引用作为标量值(例如，可以由 $a1 或 $a2[0]).

Note the difference between () and [] when assigning to arrays. Brackets [] create an anonymous array, which returns an array reference as a scalar value (for example, that can be held by $a1, or $a2[0]).

另一方面，除了改变运算符的优先级之外，括号实际上什么都不做.

Parentheses, on the other hand, do nothing at all really, except change the precedence of operators.

考虑这段代码:

my @a2 = 1, 2, 3;
print "@a2";

这将打印 1.如果你use warnings，你也会得到一个警告，比如:Useless use of a constant in void context.基本上，会发生这种情况:

This will print 1. If you use warnings, you will also get a warning such as: Useless use of a constant in void context. Basically, this happens:

my @a2 = 1;
2, 3;

因为逗号 (,) 的优先级低于等号 =.(请参阅 perldoc perlop 中的运算符优先级和关联性".)

Because commas (,) have a lower precedence than equal sign =. (See "Operator Precedence and Associativity" in perldoc perlop.)

括号简单地否定了 = 和 , 的默认优先级，并将 1,2,3 组合在一个列表中，然后传递给 @a2.

Parentheses simply negate the default precedence of = and ,, and group 1,2,3 together in a list, which is then passed to @a2.

所以，简而言之，方括号 [] 有一些魔力:它们创建匿名数组.括号，()，只是改变优先级，就像在数学中一样.

So, in short, brackets, [], have some magic in them: They create anonymous arrays. Parentheses, (), just change precedence, much like in math.

文档中有很多内容需要阅读.这里有人曾经向我展示了一个非常好的取消引用链接，但我不记得它是什么.在 perldoc perlreftut 中，您将找到有关参考的基本教程.在 perldoc perldsc 中，您会找到有关数据结构的文档(感谢 Oesor 提醒我).

There is much to read in the documentation. Someone here once showed me a very good link for dereferencing, but I don't recall what it was. In perldoc perlreftut you will find a basic tutorial on references. And in perldoc perldsc you will find documentation on data structures (thanks Oesor for reminding me).

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