in reply to Re^3: Fastest way to test membership in a constant set
in thread Fastest way to test membership in a constant set

Sorry, I saw something that I thought worked: 
use constant {
    SEC   => 0,
    MIN   => 1,
    HOUR  => 2,
    MDAY  => 3,
    MON   => 4,
    YEAR  => 5,
    WDAY  => 6,
    YDAY  => 7,
    ISDST => 8,
};

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Instead, it's just syntax sugar to declare multiple constants with one use constant statement.

I searched the doc on constant, and didn't see a good example, though

Dereferencing constant references incorrectly (such as using an array subscript on a constant hash reference, or vice versa) will be trapped at compile time.
implies that a hash constant can be created.

Here's how it's done: 

use constant Books_I_Own => 
{
    "The Wit and Wisdom of Mark Twain" => 1,
    "Attack of the Deranged Mutant Killer Monster Snow Goons" => 1,
    "Unix Power Tools" => 1,
}

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and from the debugger 
  DB<2> x Books_I_Own
0  HASH(0x1c831b8)
   'Attack of the Deranged Mutant Killer Monster Snow Goons' => 1
   'The Wit and Wisdom of Mark Twain' => 1
   'Unix Power Tools' => 1

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...so it is possible to have a constant set, if a hash is a set.

-QM
--
Quantum Mechanics: The dreams stuff is made of

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Re^5: Fastest way to test membership in a constant set
by BrowserUk (Patriarch) on Aug 10, 2007 at 15:48 UTC

    Yep. A limitation of that syntax is that, whilst the hash is a constant, it contents are not. 

    [14:26:14.28] c:\test>p1
use constant HASH=>{ a=>1, b=>2 };;
print %{ +HASH };;
a 1 b 2

HASH->{ c } = 3;;
print %{ +HASH };;
c 3 a 1 b 2

delete HASH->{ $_ } for keys %{ +HASH };;
print %{ +HASH };;

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    That can be solved using the same Hash::Util::lock_hash() method I mentioned above.

    Examine what is said, not who speaks -- Silence betokens consent -- Love the truth but pardon error.
    "Science is about questioning the status quo. Questioning authority".
    In the absence of evidence, opinion is indistinguishable from prejudice.
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