I'm having some trouble manipulating 64-bit values in Perl. `perl -V` confirms that I am running perl 5.6.1 with USE_64_BIT_INT and USE_64_BIT_ALL on Tru64 UNIX. I believe what I am seeing is that whenever I do a comparison or any arithmetic, perl with translate the value to a floating point number, losing the precision that I need. (Bit manipulation operators are okay)

As a work around, I wind up manipulating the number as two 32-bit quantities. However, this is much slower, and, IMNSHO, quite ridiculous.

Below is a script and it's output illustrating what I am describing: 

$ ./foo.pl
A) 0xfffffc0123456789 (should be 0xfffffc0123456789)
B) 0xfffffc0123456800 (should be 0xfffffc0123456789)
C) 0xfffffc0123456789 (should be 0xfffffc0123456789)
D) 0xfffffc0123456800 (should be 0xfffffc0123456788 or maybe 0xfffffc0
+123456790)
E) 0x23456789 (should be 0x23456789)
F) 0xfffffc01 (should be 0xfffffc01)
G) Comparing 0xfffffc0123456789 to 0xfffffc0123456785:
   1) 0 (should be 1)
   2) 0 (should be -1)
   3) 0 (should be 0)
$ 
$ cat foo.pl
#!/usr/local/bin/perl
$a = 0xfffffc0123456789;
printf "A) %#x (should be 0xfffffc0123456789)\n", $a;
$str = "0xfffffc0123456789";
printf "B) %#x (should be 0xfffffc0123456789)\n", hex($str);
$b = unpack "Q", (pack "L2", hex(substr($str, 10)), hex(substr($str, 2
+, 8)));
printf "C) %#x (should be 0xfffffc0123456789)\n", $b;
printf "D) %#x (should be 0xfffffc0123456788 or maybe 0xfffffc01234567
+90)\n",
    $a - 1;
printf "E) %#x (should be 0x23456789)\n", $a & 0xffffffff;
printf "F) %#x (should be 0xfffffc01)\n", $a >> 32;
$b = 0xfffffc0123456785;
printf "G) Comparing %#x to %#x:\n", $a, $b;
printf "   1) %d (should be 1)\n", $a <=> $b;
printf "   2) %d (should be -1)\n", $b <=> $a;
printf "   3) %d (should be 0)\n", $a <=> $a;
$

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In reply to Manipulating 64-bit values by rjbell4

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