I don't see the need for super efficiency for an array with 5 or 10 True/False values. Use the scalar value of grep to count the number of "zeroes". 
#!/usr/bin/perl -w
use strict;

my @results_file=(1,1,1,0,1);

print "at least one zero" if (grep{/0/}@results_file);
__END__
prints:
at least one zero

[download]
numeric compare in the grep will do the same thing: 
print "at least one zero" if (grep {$_ == 0}@results_file);

[download]
Slight modification to your code yields the following.... The grep says "tell me the number of times that '0'(zero) was seen when I looked at all elements of @results_file". Yes, there are ways to "stop counting" when the first "zero" is reached, but for just a couple of dozen things in @results_file, it just won't matter. 
#!/usr/bin/perl -w
use strict;

my @results_file=(1,1,1,0,1);

if (grep {/0/}@results_file ) #if (grep {$_ == 0}@results_file)
                              #numeric value
{
    print"\n--FAIL--";
}
else
{
     print "\n--PASS--STATUS\n";
     print "\n@results_file\n";
}

[download]

In reply to Re: Most efficient way of search elements in an array by Marshall
in thread Most efficient way of search elements in an array by sqspat

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