Hi,

I was initially going to mark this post as off-topic ... still not entirely sure whether I should have ....
The question relates specifically to perls for which both $Config{ivsize} and $Config{nvsize} are both 8. That is, perl's integer type (UV/IV) is 64-bit, and perl's floating point type (NV) is either a double or an 8-byte long double.

The aim is to determine whether a given integer value can be represented exactly as a double.
Clearly, any integer <= 9007199254740992 can be represented exactly as a double. (9007199254740992 == 2 ** 53.)
In addition to those values, however, any integer whose highest set bit and lowest set bit are separated by 51 or fewer bits is also exactly representable as a double.

Here follows my solution. The question is "Is there a better way ?".
Assume that the given arg is an integer in the range 0 .. 18446744073709551615 (with 18446744073709551615 being the largest possible UV value).
As an XSub: 
int uv_fits_double(UV arg) {

  if(arg < 9007199254740993)
    return 1;

  while(!(arg & 1)) {
    arg >>= 1;
    if(arg < 9007199254740993) return 1;
  }

  return 0;
}

[download]
And as perl sub: 
sub uv_fits_double {

  my $arg = shift;

  return 1 if $arg < 9007199254740993;

  while(!($arg & 1)) {
    $arg >>= 1;
    return 1 if $arg < 9007199254740993;
  }

  return 0;
}

[download]
It annoys me that I can't find a way to detect and shift all of the trailing zero bits off in one hit - and that I instead have to detect and shift them off one at a time.
The number of times that the "$arg < 9007199254740993" comparison is evaluated also annoys me. (Doing that evaluation inside the while loop means that the while loop will perform a maximum of 11 cycles. Without that evaluation it could perform up to 63 cycles.)

As I understand it, the only thing I need to determine is "Are there more than 51 bits between the highest set bit and the lowest set bit ?", and I do that by shifting off all trailing unset bits so that I can then determine (by examining the remaining value) whether it fits into 53 bits or not.
It feels like there ought to be a quicker, simpler way of doing it ... but I don't see one.
A perl demo of the uv_fits_double sub: 
use strict;
use warnings;
use Config;

die "This script not meant for this perl configuration"
  unless $Config{ivsize} == $Config{nvsize};

# The integer  value 2251799813685249 is
# representable exactly as a double.
# Therefore 2251799813685249 * (2 ** 10)
# is exactly representable as a double, 
# since 10 is well within the allowable 
# exponent range.
# 2251799813685249 * (2 ** 10) is also
# within the bounds of allowable integer
# values.

my $fits_d = 2305843009213694976; # 2251799813685249 * (2 ** 10)
my $no_fits_d = $fits_d + (2 ** 6);

print uv_fits_double(2251799813685249);    # fits
print uv_fits_double($fits_d);             # fits
print uv_fits_double($no_fits_d);          # doesn't fit
print uv_fits_double($fits_d + (2 ** 15)); # fits
print "\n";


sub uv_fits_double {
  my $arg = shift;

  return 1 if $arg < 9007199254740993;

  while(!($arg & 1)) {
    $arg >>= 1;
    return 1 if $arg < 9007199254740993;
  }

  return 0;
} 

__END__

Should output 1101
The 4th value fits, even though it's greater
than the 3rd value (which doesn't fit).

[download]
Cheers,
Rob

In reply to Detecting whether UV fits into an NV by syphilis

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