how do I find if at least one of many keys in a list is in a hash (quick, fast, and simple)

Dunno if this meets *all* of your criteria, but I find this simple enough:

my @search_keys = qw(one two three);
my %hash = ( four=>4,
         five=>5,
        six=>6);
my $foundit =0;
foreach (@search_keys) {
    if (exists $hash{$_}) {
        $foundit =1;
        print "found an entry for $_ in %hash!\n";
        }
}
[download]

Philosophy can be made out of anything. Or less -- Jerry A. Fodor

Given that the idea is to find out if any of the keys exist in the hash, it would be more efficient to call last and exit the loop once you've found one.

--
<http://www.dave.org.uk>

"Perl makes the fun jobs fun
and the boring jobs bearable" - me

@keys = qw(one two five);
unshift @keys,'four';
print "@keys\n";
if (@hash{@keys}) {
    print "yep\n";
} else {
    print "nope\n";
}
 
@keys = qw(one two five);
push @keys, 'two';
print "@keys\n";
if (@hash{@keys}) {
    print "yep\n";
} else {
    print "nope\n";
}
[download]

four one two five
yep
one two five two
nope 
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if ( grep{ exists( $hash{$_} ) } @keys ) { 
    print "yep\n";
} else {
    print "nope\n";
}
[download]

Ayup -- it's the last key that matters, just as if the hash slice were being treated as a list -- evaluate a list in a scalar context and it returns the last value.

There's nothing in the docs that I could find that says that a hash slice would be interpreted as a list rather than a proper array, but it does apear to be the case.

Any one further investigation, it seems that my original solution suffered from a fundamental problem.

my %hash = (one => 1, two => 2, three => 3);
my @keys = qw(four five six);
my @slice = @hash{@keys};
[download]

@slice is now contains four undef elements, where I thought it would be an empty list (don't know why I thought that - it was very early in the morning.

If the hash slice was interpreted as an array, it would therefore always be true and therefore doesn't solve ChOas' original problem. That would be better addressed using something like:

if (grep { exists %hash{$_} } @keys) {
  print "yep\n";
} else {
  print "nope\n";
}
[download]

--
<http://www.dave.org.uk>

"Perl makes the fun jobs fun
and the boring jobs bearable" - me

Well, we always go around about this, because the perldata does clearly say:

           @days               # ($days[0], $days[1],... $days[n])
           @days[3,4,5]        # same as @days[3..5]
           @days{'a','c'}      # same as ($days{'a'},$days{'c'})
[download]

See the "same as"? That's the operative words. To make it the same, it has to return the last element of the list in a scalar context. So yes, the behavior is documented, and derivable from the docs.

-- Randal L. Schwartz, Perl hacker