list 如何删除数组的前五个元素？
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How do I remove the first five elements of an array?
提问by nebulus
@array = qw(one two three four five six seven eight);
<Some command here>
print @array;
回答by friedo
Here are a few ways, in increasing order of dumbness:
这里有几种方法，按愚蠢程度递增：
Using a slice:
使用切片：
@array = @array[ 5 .. $#array ];
Using splice:
使用splice：
splice @array, 0, 5;
Using shift:
使用shift：
shift @array for 1..5;
Using grep:
使用grep：
my $cnt = 0;
@array = grep { ++$cnt > 5 } @array;
Using map:
使用map：
my $cnt = 0;
@array = map { ++$cnt < 5 ? ( ) : $_ } @array;
I'm sure far better hackers than I can come up with even dumber ways. :)
我敢肯定，比我想出的更愚蠢的方法要好得多的黑客。:)
回答by Anomie
splice @array, 0, 5;will do it.
splice @array, 0, 5;会做的。
回答by Joel Berger
As a comment to friedo's answer and to demonstrate cool new declaration state, here it is using grep, which friedo's mapemulates.
作为对 Friedo 的回答的评论并展示很酷的新声明state，这里使用的grep是 Friedo 的map模拟。
#!/usr/bin/perl
use strict;
use warnings;
use feature 'state';
my @array = qw(one two three four five six seven eight);
my @new_array = grep {state $count; ++$count > 5} @array;
print "$_\n" for @new_array;
回答by cyber-guard
I just realized you only need the last string, so no need to loop
我刚刚意识到你只需要最后一个字符串，所以不需要循环
my $_ = "@array"; s|(?:.*?\s){5}||;say;
Btw this is probably the least efficient way to do it, just having fun :)
顺便说一句，这可能是效率最低的方法，只是玩得开心:)
