Re: Most efficient way of search elements in an array

Use a for loop to cycle through your array elements, and break out of the loop as soon as 0 is found using last:

use strict;
use warnings;

my @results_file = (1,1,1,0,1);
for (@results_file) {
    if (/0/) {
        print "\n--FAIL--\n";
        last;
    }
    else {
        print "\n--PASS--STATUS\n";
        print "\n$_\n";
    }
}

__END__

--PASS--STATUS

1

--PASS--STATUS

1

--PASS--STATUS

1

--FAIL--
[download]

Update: Or maybe this is the output you were looking for...

use strict;
use warnings;

my @results_file = (1,1,1,0,1);
my $pass = 1;
for (@results_file) {
    if (/0/) {
        print "\n--FAIL--\n";
        $pass = 0;
        last;
    }
}
if ($pass) {
    print "\n--PASS--STATUS\n";
    print "\n@results_file\n";
}

__END__

--FAIL--
[download]

Comment on Re: Most efficient way of search elements in an array Select or Download Code

Replies are listed 'Best First'.
Re^2: Most efficient way of search elements in an array by sqspat (Acolyte) on Sep 15, 2009 at 15:16 UTC
This does it alright ... except I just want one PASS to be printed if the script determines that all the array elements are 1	[reply]
Re^3: Most efficient way of search elements in an array by toolic (Bishop) on Sep 15, 2009 at 15:23 UTC
Then you want my 2nd example ("Update"). Also note that my code is equivalent to the pre-5.10 solution offered by Fletch. It wasn't obvious to me from the documentation for List::MoreUtils, until I looked at the source code for `any`: it does break out of the loop as soon as it finds the first match.	[reply] [d/l]