使用條件項目生成列表

Question

是否可以使用條件項目創建數組？

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

這樣"no-op"是功能等作品，如果$condition是假的，然後我得到的名單(1, 3)但如果$condition是真的，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("."); 

在某些情況下，我想包括線

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

而在其他情況下，我不希望排除目錄_private.d ..

Answer 1

您可以使用空列表()跳過第二個元素，

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

Answer 2

一般來說，你可以使用

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

在上下文中得到一些可讀性的收益，這將是

my @rules; 
push @rules, $rule->new->name('*.cfg')->prune->discard; 
push @rules, $rule->directory->name("_private.d")->prune->discard; 
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(".");