[英]Range operator; flip but prevent flop, and make immediate flip
perl -wle 'print join " ", grep /3/ .. undef(), 1..10'
outputs 3 4 5 6 7 8 9 10
输出
3 4 5 6 7 8 9 10
Q1
: Is there better way than undef
to prevent flop? Q1
:有没有比undef
更好的方法来阻止翻牌?
Q2
: How to force left part of range operator to unconditional true
(ie. true .. /7/
)? Q2
:如何强制范围运算符的左边部分为无条件true
(即true .. /7/
)?
UPDATE: 更新:
perl -wE 'say join " ", grep { ((/7/ .. undef)||1) ==1 } 1..10'
could be used as true .. /7/
replacement. 可以用作
true .. /7/
替换。
Any false expression that isn't constant-folded to a number will do. 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'
Without a flip-flop. 没有触发器。
perl -wE'my $ok; say join " ", grep $ok ||= $_==3, 1..10'
If you want the boolean opposite of something, use negation! 如果你想要布尔值相反的东西,使用否定!
perl -wE'say join " ", grep !($_==8 .. undef), 1..10'
Without a flip-flop. 没有触发器。
perl -wE'my $done; say join " ", grep !($done ||= $_==8), 1..10'
Ok, so I changed 7
to 8
. 好的,所以我把
7
改为8
。 To actually match on 7
, 要实际匹配
7
,
perl -wE'my $last; say join " ", grep { my $x = ($_==7 .. undef); !$x || $x == 1 } 1..10'
Without a flip-flop. 没有触发器。
perl -wE'my $done; say join " ", grep { my $rv = $done; $done ||= $_==7; !$rv } 1..10'
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