Welcome BrianP,

On a Linux 32 bit system, Perl still allows 2**53 integers if you compile Perl for 64 bit integers: (Note: If Perl is a 64 bit version, then the problem goes away!) 

rperl -e 'use integer;$i=((2**17)**3);$m=2**53;print "$i\n$m\n";'
2251799813685248
9007199254740992

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As you can see the maximum integer is larger than what you need. So you can generate integers directly by multiplying the 3 16 bit
<UPDATE>
To make sure nobody just multiplies the RGB together please remember that to represent the number 307 in decimal use: 
perl -e '$n = (3*(10**2)) + (0*(10**1)) + (9*(10**0)); print "$n\n";'

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So the formula is ($B is the base): 
perl -e '$B = 10; $n = (3*($B**2)) + (0*($B**1)) + (9*($B**0)); print 
+"$n\n";'

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So to generate 48 bit RGB integers you need: 
perl -e '$B = 2**16; $n = ($R*($B**2)) + ($G*($B**1)) + ($B*($B**0)); 
+print "$n\n";'

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Clarification complete!
</UPDATE>
values and then use that value as a hash key. Then each time you add a count just do: 
$hash{$key}++;             ## $key is 48 bit RGB as a single integer
                           ## The value $hash{$key} is sum of the time
+s found

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Regards...Ed

"Well done is better than well said." - Benjamin Franklin

In reply to Re: Perl Hashes in C? (Updated) by flexvault
in thread Perl Hashes in C? by BrianP

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