hmm maybe simpler than we thought!
DB<25> sub first (&@) { my $code=shift; grep {return $_ if &$code} @
+_ }
DB<26> first { $_ >5 } 1..10
=> 6
DB<27> first { $_ >10 } 1..10
DB<28> sub any (&@) { my $code=shift; grep {return 1 if &$code} @_ ;
+ ()}
DB<29> any { $_ >10 } 1..10
=> ""
DB<30> any { $_ >5 } 1..10
=> 1
Can you spot problems?
update
variation of first for finding position
DB<40> sub position (&@) { my $code=shift; my $idx=-1; grep {++$idx;
+ return $idx if &$code} @_ ; ()}
DB<41> position { !$_ } 1..10,undef,11..20
=> 10
update
changed "" to () for false.
update
even easier
DB<53> sub first (&@) { my $code=shift; &$code and return $_ for @_
+; ()}
DB<54> sub any (&@) { my $code=shift; &$code and return 1 for @_ ; (
+)}
Cheers Rolf
( addicted to the Perl Programming Language)
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