Re^3: Highest scalar = ???

I suppose that depends on your definition of "integer". I was never much of a math student, so I use the comp. sci definition. So does Perl, sometimes:

   $x = 2**(52+1);
   for (1 .. $x) {
     # not gonna get here!
   }
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Produces:

Range iterator outside integer range at foo.pl line 2.

Guess that's not an integer after all!

-sam

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Re^4: Highest scalar = ??? by ikegami (Patriarch) on Oct 01, 2007 at 20:27 UTC
I was using your own definition. You said `~0` was the highest integer that can be represented precisely without using Math::BigInt. That's wrong, and I explained why. Your original post if still wrong by your new definition. If you want to the highest integer that can be used in a range, both your answer of `~0` and your suggestion to use Math::BigInt for larger integers than `~0` are wrong!! I have already mentioned in this thread that ranges require signed integers, so `~0 >> 1` (2147483647) would be the answer. Update: I originalyl had `~-(~0+1)` instead of `~0 >> 1`, but that won't work if the system's floats have a smaller range than the integers.	[reply] [d/l] [select]

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Re^4: Highest scalar = ???
by ikegami (Patriarch) on Oct 01, 2007 at 20:27 UTC

I was using your own definition. You said ~0 was the highest integer that can be represented precisely without using Math::BigInt. That's wrong, and I explained why.

Your original post if still wrong by your new definition. If you want to the highest integer that can be used in a range, both your answer of ~0 and your suggestion to use Math::BigInt for larger integers than ~0 are wrong!!

I have already mentioned in this thread that ranges require signed integers, so ~0 >> 1 (2147483647) would be the answer.

Update: I originalyl had ~-(~0+1) instead of ~0 >> 1, but that won't work if the system's floats have a smaller range than the integers.

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