我不明白这小段代码是怎么回事。
我已经把关于它的输出和评论放在中间,在产生它的打印语句之后
my @regions = (
0,
1,
2,
[ 0, new Point(3,1), new Point(3,2), new Point(3,3) ],
4,
5 );
$reg1 = 3;
print "1: $regions[$reg1] n";
@reg1 = @{regions[$reg1]};
print "2: $reg1[0]n";
print "3: $reg1[0][1]n";
print "4: ", Point::stringPoint($reg1[0][1]), "n";
# HERE IS THE OUTPUT from the above print statements, with comments about my confusion appended
1: ARRAY(0xe8b0e0) # ok - element 3 of @regions is an array, as expected.
2: ARRAY(0xe8b0e0) # It appears to be an array of one element, which is itself. ???
3: Point=HASH(0xfda5e0) # We have go 'down' one more level to get what we want - that makes no sense
4: 3,1 # Yes, there it is
package Point;
sub new {
my $class = shift;
my $self = {
x => shift,
y => shift
};
bless $self, $class;
return $self;
}
sub stringPoint
{
my $p = shift;
return "$p->{x},$p->{y}";
}
" Code related to new question (with output) " ;
我真正的问题是:
如何直接方便地使用另一个数组中的数组(而不是它的副本(
唯一的方法是(总是(取消引用引用吗
例如下面的两个不起作用的例子。
以下是我尝试过的:
my $ref1 = @{$regions[3]};
@{$ref1}[2] = new Point(4, 5); # changes original array
print1Region(3, $ref1);
# OUTPUT = (3,1) (4,5) (3,3)
my @arr1 = @{$ref1};
$arr1[1] = new Point(2,6); # does not
print1Region(3, $ref1);
# OUTPUT = (3,1) (4,5) (3,3)
$ref1[0] = new Point(1,4); # does not
print1Region(3, $ref1);
# OUTPUT = (3,1) (4,5) (3,3)
@{regions[$reg1]}
是一种奇怪且未记录的@regions[$reg1]
编写方式。(这是一种用于双引号字符串文字的语法文档。(
@regions[$reg1]
是一个只有一个元素的数组切片,这是一种奇怪的$regions[$reg1]
写入方式。
因此,您并没有得到$regions[$reg1]
引用的数组的第一个元素;您只是得到了@regions
的第一个元素。
让我们看看
my $ref1 = @{$regions[3]};
@{$ref1}[2] = new Point(4, 5);
问题#1
@
"取消",因此
my $ref1 = @{$regions[3]};
只是编写的一种复杂方式
my $ref1 = $regions[3];
(好吧,这不太正确,因为前者会自动激活,但这与这里无关。(
问题#2
再次使用一个元素的数组切片。务必使用use strict; use warnings;
并注意警告!
@{$ref1}[2] = new Point(4, 5);
应该是
${$ref1}[2] = new Point(4, 5);
使用"箭头符号"写得更干净。
$ref1->[2] = new Point(4, 5);
问题#3
最后,不要使用间接方法调用。它们会引发问题。
$ref1->[2] = new Point(4, 5);
应该是
$ref1->[2] = Point->new(4, 5);
结论
my $ref1 = @{$regions[3]};
@{$ref1}[2] = new Point(4, 5);
应该写成
my $ref1 = $regions[3];
$ref1->[2] = Point->new(4, 5);
如果没有变量,这将是
$regions[3]->[2] = Point->new(4, 5);
或者只是
$regions[3][2] = Point->new(4, 5);