范围算子; 翻转,但防止翻牌,并立即翻转

Question

perl -wle 'print join " ", grep /3/ .. undef(), 1..10'

输出 3 4 5 6 7 8 9 10

Q1:有没有比undef防止翻牌更好的方法？

Q2:如何强制范围运算符的左侧部分无条件true(即.true .. /7/)？

更新:

perl -wE 'say join " ", grep { ((/7/ .. undef)||1) ==1 } 1..10'

可以用作true .. /7/替代品.

Answer 1

1. 任何非常数折叠为数字的虚假表达都可以.

   perl -wE'say join " ", grep $_==3 .. undef, 1..10'
   
   perl -wE'say join " ", grep $_==3 .. do{0}, 1..10'
   
   perl -wE'say our $FALSE; say join " ", grep $_==3 .. $FALSE, 1..10'
   

   没有触发器.

   perl -wE'my $ok; say join " ", grep $ok ||= $_==3, 1..10'
   

2. 如果你想要布尔值相反的东西,使用否定!

   perl -wE'say join " ", grep !($_==8 .. undef), 1..10'
   

   没有触发器.

   perl -wE'my $done; say join " ", grep !($done ||= $_==8), 1..10'
   

   好的,所以我换7了8.要实际匹配7,

   perl -wE'my $last; say join " ", grep { my $x = ($_==7 .. undef); !$x || $x == 1 } 1..10'
   

   没有触发器.

   perl -wE'my $done; say join " ", grep { my $rv = $done; $done ||= $_==7; !$rv } 1..10'