From perldata: "If you evaluate a hash in scalar context, it returns false if the hash is empty. If there are any key/value pairs, it returns true; more precisely, the value returned is a string consisting of the number of used buckets and the number of allocated buckets, separated by a slash."..

Am I right to assume that if the hash is non-empty if (%hash) and scalar(%hash) are simply returning a stored number of buckets (no loops to count buckets)? If so, that would be exactly want I wanted - O(1). I missed that paragraph. Thanks.

Best, beth

Update: just to be clear - this is meant to be an internals question, not a "trust the docs" question. perlguts doesn't go that deep into the internals of hash implementation.

That's what I expect from common sense, yes.

A quick benchmark seems to confirm this; asking an empty hash is much faster than asking a filled one, but the number of keys in the full hash seems to have no significant influence on the speed:

use strict;
use warnings;
use Benchmark qw(cmpthese);

for my $len (2..6) {
    our (%full, %empty);

    my $a = 'A';
    $full{$a++} = $a while length($a) < $len;

    print "\nLength $len with ", scalar(%full), "items\n";
    cmpthese -1, {
        full    => sub {die unless %full},
        empty    => sub {die if %empty},
    }
}
__END__
Length 2 with 22/32items
           Rate  full empty
full  1286220/s    --  -77%
empty 5481192/s  326%    --

Length 3 with 510/1024items
           Rate  full empty
full  1052183/s    --  -82%
empty 5748771/s  446%    --

Length 4 with 13961/32768items
           Rate  full empty
full   998734/s    --  -84%
empty 6056132/s  506%    --

Length 5 with 310721/524288items
           Rate  full empty
full   963764/s    --  -83%
empty 5681139/s  489%    --

Length 6 with 8655839/16777216items
           Rate  full empty
full   849541/s    --  -86%
empty 6124859/s  621%    --
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Note that the number of hash items grows exponentially, while the number of iterations decreases linearly at best.

Devel::Peek and a quick perusal of the sources seems back up the idea that its a lookup

C:\> perl -MDevel::Peek -e"Dump { }
SV = RV(0x183cfc8) at 0x226074
  REFCNT = 1
  FLAGS = (TEMP,ROK)
  RV = 0x225f84
  SV = PVHV(0x22b504) at 0x225f84
    REFCNT = 1
    FLAGS = (SHAREKEYS)
    IV = 0         <<<------------------------------------------
    NV = 0
    ARRAY = 0x0
    KEYS = 0       <<<------------------------------------------
    FILL = 0
    MAX = 7
    RITER = -1
    EITER = 0x0

C:\>perl -MDevel::Peek -e"Dump { 1..10 }
SV = RV(0x183cfc8) at 0x182f4f4
  REFCNT = 1
  FLAGS = (TEMP,ROK)
  RV = 0x225f94
  SV = PVHV(0x22b514) at 0x225f94
    REFCNT = 1
    FLAGS = (SHAREKEYS)
    IV = 5         <<<------------------------------------------
    NV = 0
    ARRAY = 0x1825554  (0:3, 1:5)
    hash quality = 150.0%
    KEYS = 5       <<<------------------------------------------
    FILL = 5
    MAX = 7
    RITER = -1
    EITER = 0x0
    Elt "1" HASH = 0x806b80c9
    SV = IV(0x1828ce0) at 0x226084
      REFCNT = 1
      FLAGS = (IOK,pIOK)
      IV = 2
    Elt "3" HASH = 0xa400c7f3
    SV = IV(0x1828d04) at 0x2260b4
      REFCNT = 1
      FLAGS = (IOK,pIOK)
      IV = 4
    Elt "7" HASH = 0xecc9d984
    SV = IV(0x1828d14) at 0x2260f0
      REFCNT = 1
      FLAGS = (IOK,pIOK)
      IV = 8
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Thanks for the links to Perl source code and the dump.

Key counts being stored would be consistent with BrowserUk's observations that there seems to be little difference between scalar keys on an empty and full hash. But scalar(%h) is returning counts of used and allocated buckets, not key counts.

Given moritz's observations of a significant difference in empty and non-empty hashes for %h and the lack of an obvious dump field to store bucket counts in Devel::Peek output, I would think that %h is doing some sort of looping and is O(number of buckets).

Best, beth

No, the documentation is all lies :)

C:\>perl -e"die scalar %hash
0 at -e line 1.

C:\>perl -e"die scalar %ENV
26/64 at -e line 1.
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