Range operator; flip but prevent flop, and make immediate flip

perl

Solution

Any false expression that isn't constant-folded to a number will do.

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.

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

If you want the boolean opposite of something, use negation!

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

Without a flip-flop.

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

Ok, so I changed `7` to `8`. To actually match on `7`,

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

Without a flip-flop.

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

Problem

``` perl -wle 'print join " ", grep /3/ .. undef(), 1..10' ``` outputs `3 4 5 6 7 8 9 10` `Q1`: Is there better way than `undef` to prevent flop? `Q2`: How to force left part of range operator to unconditional `true` (ie. `true .. /7/`)? UPDATE: ``` perl -wE 'say join " ", grep { ((/7/ .. undef)||1) ==1 } 1..10' ``` could be used as `true .. /7/` replacement.

Original source