in reply to Largest integer in 64-bit perl

Your value is 2**53-1, not because it's the biggest 32bit Integer value, but because Perl automatically swaps to floating point with a 53 bit mantissa if the 32 bits are exceeded (see Double-precision floating-point format and perlnumber )

The max value for 64 bit representation will be about 2**63-1, because you need one bit for the sign in signed integers.

This has already been discussed many times.

NB: all values probably off by one, (UPDATED -1 because one slot will be occupied by zero.)

I'm not going to look up the correct values for you, if you're too lazy¹ to supersearch or Google, so am I.

FWIW: Your linear search is not terribly clever, use a binary search and you'll have finished before pronouncing the s in stupid.

Cheers Rolf
(addicted to the Perl Programming Language :)
see Wikisyntax for the Monastery

¹) quote: > "stored in IEEE-754 format internally. Whatever that means."

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Re^2: Largest integer in 64-bit perl
by ikegami (Patriarch) on May 18, 2025 at 04:07 UTC

    How would you use a binary search to find the first gap?

[reply]
      I would try to test the OP's problem that predecessor or successor are identical in numerical comparison.

      use v5.12;
use warnings;

#say "Mode: ", my $mode = $ARGV[0] // 0;

use Config; say $Config{ivsize}; say $Config{nvsize};

sub test {
  my ($mode) = @_;
  say "**** Mode: $mode";
  my $now=1;
  for my $x ( 1.. 65) {
    $now =
      $mode == 0 ? 2 * $now    :
      $mode == 1 ? 2 ** $x     :
      $mode == 2 ? 2 ** $x  -1 :
      die "Mode=$mode not implemented yet";

    if ( $now == $now-1 or $now == $now+1 ) {
      print "Problem at $now = 2**$x\n";
      printf "%22u %22u %22u\n",$now-1,$now,$now+1;
      last;
    }
  }
}

test($_) for 0..2;

      [download]

      this will break at 2**64 when I stay purely integer

      NB: I didn't implement the full binary search, since there is no backtracking yet

      But I'm getting weird conversion results when leaving the realm of integer (see the "modes"), these are most likely side-effects of starting with a float (or bugs). Probably better to use Hex-strings. 

      perl /home/lanx/perl/pm/integer-gap.pl
8
8
**** Mode: 0
Problem at 1.84467440737096e+19 = 2**64
  18446744073709551615   18446744073709551615   18446744073709551615
**** Mode: 1
Problem at 9.00719925474099e+15 = 2**53
      9007199254740991       9007199254740992       9007199254740993
**** Mode: 2
Problem at 4.61168601842739e+18 = 2**62
   4611686018427387904    4611686018427387904    4611686018427387904

      [download]

      YMMV...

      Cheers Rolf
      (addicted to the Perl Programming Language :)
      see Wikisyntax for the Monastery

      Update

      In hindsight, the cleanest approach would be to only rely on addition and substraction of integers. No multiplications or exponentions (which might be implemented as approximation)

      Since the last safe interval is guaranteed to be an integer too.

[reply]
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        I would try to test the OP's problem that predecessor or successor are identical in numerical comparison.

        I haven't looked too closely, but I suspect that the fact that the power operator (**) always returns an NV might be causing problems.
        You might be better off using the << operator: 
        D:\>perl -MDevel::Peek -we "Dump (2**62); Dump ((2**62) + 1); Dump ((2
+**62) - 1);"
SV = NV(0x1d26f8d3f20) at 0x1d26f8d3f38
  REFCNT = 1
  FLAGS = (PADTMP,NOK,READONLY,PROTECT,pNOK)
  NV = 4.6116860184273879e+18
SV = NV(0x1d26f8dbb88) at 0x1d26f8dbba0
  REFCNT = 1
  FLAGS = (PADTMP,NOK,READONLY,PROTECT,pNOK)
  NV = 4.6116860184273879e+18
SV = NV(0x1d26f8d4ac0) at 0x1d26f8d4ad8
  REFCNT = 1
  FLAGS = (PADTMP,NOK,READONLY,PROTECT,pNOK)
  NV = 4.6116860184273879e+18

D:\>perl -MDevel::Peek -we "Dump (1<<62); Dump ((1<<62) + 1); Dump ((1
+<<62) - 1);"
SV = IV(0x132ea044988) at 0x132ea044998
  REFCNT = 1
  FLAGS = (PADTMP,IOK,READONLY,PROTECT,pIOK)
  IV = 4611686018427387904
SV = IV(0x132ea0449e8) at 0x132ea0449f8
  REFCNT = 1
  FLAGS = (PADTMP,IOK,READONLY,PROTECT,pIOK)
  IV = 4611686018427387905
SV = IV(0x132ea0442c8) at 0x132ea0442d8
  REFCNT = 1
  FLAGS = (PADTMP,IOK,READONLY,PROTECT,pIOK)
  IV = 4611686018427387903

        [download]
        The first one-liner results in 3 identical values; the second one-liner results in 3 different values.
        It's all just part of the fun we have when we use the utterly insane perl configuration where IV precision is greater than NV precision ;-)

        Cheers,
        Rob
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