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++;
}

[download]
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

[download]

In reply to Re^2: Highest scalar = ??? by ikegami
in thread Highest scalar = ??? by Anonymous Monk

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