请问这句话是啥意思啊 push @d, [ @row ];

Question

是把row这个数组的地址赋给数组d吗

Answer 1

不是  是把row的内容复制一遍  然后再把新地址push到d里面
push @d, \@row  这个才是

Answer 2

回复 1# hbhswxy
push @row的引用 into @d as the last member of array.

Answer 3

1. use strict;
3. my @list = (1,2,3,4,5,6);
4. my @arr;
6. push @arr,[@list];
7. $list[5] = 7;
8. print $arr[0]->[5],"\n";
10. push @arr,\@list;
11. $list[5] = 8;
12. print $arr[1]->[5],"\n";
13. #输出是6和8

复制代码[@list]和\@list是不同的，[@list]是新构造了一个数组引用，和原来的数组在内存中是不同地址。而\@list是直接取引用，该引用指向数组@list，所以是同一块内存地址。

Answer 4

是不是说更改[@list]中的内容不会影响@list中的内容； 而更改\@list中的内容则会更改@list中的内容。
谢谢哦。

Answer 5

本帖最后由 Perl_Er 于 2011-04-14 13:55 编辑

Taking References
An array is a defined area of storage for list values, so we can generate a reference to it with the
backslash operator:
$arrayref = \@array;
This produces a reference through which the original array can be accessed and manipulated.
Alternatively, we can make a copy of an array and assign that to a reference by using the array
reference constructor (also known as the anonymous array constructor) [...]:
$copyofarray = [@array];
Both methods give us a reference to an anonymous array that we can assign to, delete from, and modify.
The distinction between the two is important, because one will produce a reference that points to the
original array, and so can be used to pass it to subroutines for manipulations on the original data,
whereas the other will create a copy that can be modified separately.

1.      1  #!/usr/bin/perl
2.      2  use strict;
3.      3  use warnings;
4.      4  use Data::Dumper;
6.      5  my @ary = qw/aa bb cc/;
7.      6  my @rary = [ @ary ];
10.      7  print Dumper \@ary;
11.      8  print "\n";
12.      9  print Dumper \@rary;
14.     10  $rary[0]->[0] = 100;
16.     11  print Dumper \@ary;
17.     12  print Dumper \@rary;
18. zhuj2 ( dialxtap01 ) @ /home/zhuj2/perl
19. # ./ref.pl
20. $VAR1 = [
21.           'aa',
22.           'bb',
23.           'cc'
24.         ];
26. $VAR1 = [
27.           [
28.             'aa',
29.             'bb',
30.             'cc'
31.           ]
32.         ];
33. $VAR1 = [
34.           'aa',
35.           'bb',
36.           'cc'
37.         ];
38. $VAR1 = [
39.           [
40.             100,
41.             'bb',
42.             'cc'
43.           ]
44.         ];

复制代码

Answer 6

回复 4# wxlfh

    对，利用@row 和[ ]gu构造匿名arry的引用

Answer 7

@d 里放的是row的引用

Answer 8

匿名引用的 用法了。学习了。