Re^2: Highest scalar = ???

That's not true. Floating points can represent a large range of integers exactly, and that range for a double is bigger than the range of an unsigned 32 bit int.

Type                              Min       Max
--------------------------------  --------  -----------
signed n-bit int                  -2^(n-1)  (2^(n-1))-1
unsigned n-bit int                0         (2^n)-1
IEEE float with n-bit mantissa    -2^(n+1)  2^(n+1)
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Double has n=52, so 4294967295 isn't the maximum that can be held without Math::BigInt.

Type                              Min                Max
--------------------------------  -----------------  ----------------
signed 32-bit integer                   -2147483648        2147483647
unsigned 32-bit integer                           0        4294967295
IEEE float with 52-bit mantissa   -9007199254740992  9007199254740992
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Demo:

$x = 2**(52+1);

# For doubles, format "%.17e" will tell you if
# the number is stored with no loss of precision
# because log10(2**(n+1)) < 17.
printf("%.17e\n\n", $x);

$x -= 5;
for (1..10) {
   printf("%f\n", $x);
   $x++;
}
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9.00719925474099200e+015  <-- That's exact.

9007199254740987.000000
9007199254740988.000000
9007199254740989.000000
9007199254740990.000000
9007199254740991.000000
9007199254740992.000000
9007199254740992.000000   <-- Least significant
9007199254740992.000000       bit dropped.
9007199254740992.000000
9007199254740992.000000
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Re^3: Highest scalar = ??? by samtregar (Abbot) on Oct 01, 2007 at 20:00 UTC
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! }` [download] Produces: `Range iterator outside integer range at foo.pl line 2.` Guess that's not an integer after all! -sam	[reply] [d/l] [select]
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]