On Sep 22, Bakken, Luke said:
>$/ = ''; print for grep [ chomp, s/\n\s+/ /g, s/\z/\n\n/ ], <>;
>
>I've never seen grep used with [ ] brackets before - it works the same
>with { }:
>
>print for grep { chomp, s/\n\s+/ /g, s/\z/\n\n/ } <>;
Not really. Here's an example:
print grep [ rand(2) < 1 ], 1 .. 10;
vs.
print grep { rand(2) < 1 } 1 .. 10;
The first code prints all ten numbers. The second code prints roughly
half of them. Why? Because [...] constructs a reference to an anonymous
array, and any reference is true in a boolean context. Therefore, the
code that makes up the single element in the array (rand(2) < 1) is
executed, and the reference is returned to grep() for true/false. Since
all references are true, all numbers 1 through 10 pass through the filter.
We just have the benefit of executing that code on all 10 elements, and it
doesn't matter whether rand(2) is less than 1 or not. In the SECOND code,
however, rand(2) < 1 IS the true/false test. Basically, if you want to
write a "filter" that executes code on each element and returns the
element (possibly modified) regardless of what the return value of the
filter would be, you can do:
grep [ ... ], LIST
# or perhaps
grep [ do {...} ], LIST
# or
grep { ...; 1; } LIST
--
Jeff "japhy" Pinyan [EMAIL PROTECTED] http://www.pobox.com/~japhy/
RPI Acacia brother #734 http://www.perlmonks.org/ http://www.cpan.org/
<stu> what does y/// stand for? <tenderpuss> why, yansliterate of course.
[ I'm looking for programming work. If you like my work, let me know. ]
--
To unsubscribe, e-mail: [EMAIL PROTECTED]
For additional commands, e-mail: [EMAIL PROTECTED]
