Perl:数组引用与匿名数组 [英] Perl: array reference versus anonymous array

问题描述

这可能是一个愚蠢的问题...以下代码分别输出@arrayref和@arraycont的内容.注意它们之间的区别以及它们的值分配方式.我知道匿名数组的作用，但是有人可以解释为什么存在差异吗?

This may be a silly question... The following code outputs the contents of @arrayref and @arraycont respectively. Note the difference between them and the way the values of them are assigned. I know what the anonymous array does, but can anybody explain why there is a difference?

非常感谢您.

@arrayref = ();
@array = qw(1 2 3 4);
$arrayref[0] = \@array;
@array = qw(5 6 7 8);
$arrayref[1] = \@array;
print join "\t", @{$arrayref[0]}, "\n";
print join "\t", @{$arrayref[1]}, "\n";

@arraycont = ();
@array = qw(1 2 3 4);
$arraycont[0] = [@array];
@array = qw(5 6 7 8);
$arraycont[1] = [@array];
print join "\t", @{$arraycont[0]}, "\n";
print join "\t", @{$arraycont[1]}, "\n";

输出

5   6   7   8   
5   6   7   8   
1   2   3   4   
5   6   7   8   

推荐答案

这将创建数组的浅副本:

$arraycont[0] = [@array];

这只是创建了对它的引用:

Whereas this just creates a reference to it:

$arrayref[0] = \@array;

自从您以后修改数组以来:

Since you later modify the array:

@array = qw(5 6 7 8);

arrayref仍指向内存中的相同数组位置，因此在print语句中取消引用时，它会打印当前数组值5 6 7 8.

arrayref still points to the same array location in memory, and so when dereferenced in the print statements it prints the current array values 5 6 7 8.

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