生成包含条件项的列表

Question

是否可以创建包含条件项的数组？

my @a = (1, ($condition) ? 2 : "no-op", 3);

这样 "no-op" 是这样工作的函数，如果 $condition 为假，那么我得到列表 (1, 3) 但如果 $condition 为真，我得到 (1, 2, 3) ?

背景：

use strict;
use warnings;
use File::Find::Rule;

my $rule = File::Find::Rule->new();
$rule->or(
    $rule->new->name('*.cfg')->prune->discard,
    $rule->directory->name("_private.d")->prune->discard,
    $rule->new->name('*.t')->prune->discard,
    $rule->new->name('*.bak')->prune->discard,
    $rule->new->name('.*.bak')->prune->discard,
    $rule->new->name('.#*')->prune->discard,
);

my @files = $rule->in(".");

在某些情况下，我想包含行

$rule->directory->name("_private.d")->prune->discard

在其他情况下，我不想排除目录 _private.d..

Answer 1

您可以使用空列表 () 跳过第二个元素，

my @a = (1, ($condition ? 2 : ()), 3);

Answer 2

一般来说，您可以使用

提高可读性

my @a;
push @a, 1;
push @a, 2 if $condition;
push @a, 3;

在上下文中，那将是

my @rules;
push @rules, $rule->new->name('*.cfg')->prune->discard;
push @rules, $rule->directory->name("_private.d")->prune->discard if $condition;
push @rules, $rule->new->name('*.t')->prune->discard;
push @rules, $rule->new->name('*.bak')->prune->discard;
push @rules, $rule->new->name('.*.bak')->prune->discard;
push @rules, $rule->new->name('.#*')->prune->discard;

my $rule = File::Find::Rule->new()->or(@rules);
my @files = $rule->in(".");

生成包含条件项的列表

Generate list with conditional items

perl