Ruby max integer

Question

Hello, I need to be able to determine a systems maximum integer in Ruby. Anybody know how, or if it's possible?

Answer 1

In ruby Fixnums are automatically converted to Bignums.

To find the highest possible Fixnum you could do something like this:

class Fixnum
 N_BYTES = [42].pack('i').size
 N_BITS = N_BYTES * 8
 MAX = 2 ** (N_BITS - 2) - 1
 MIN = -MAX - 1
end
p(Fixnum::MAX)

Shamelessly ripped from a ruby-talk discussion. Look there for more details.

Answer 2

Ruby automatically converts integers to a large integer class when they overflow, so there's (practically) no limit to how big they can be.

If you are looking for the machine's size, i.e. 64- or 32-bit, I found this trick at ruby-forum.com (http://www.ruby-forum.com/topic/177435):

machine_bytes = ['foo'].pack('p').size
machine_bits = machine_bytes * 8
machine_max_signed = 2**(machine_bits-1) - 1
machine_max_unsigned = 2**machine_bits - 1

Edit: If you are looking for the size of Fixnum objects (integers small enough to store in a single machine word), you can call 0.size to get the number of bytes. I would guess it should be 4 on 32-bit builds, but I can't test that right now. Also, the largest Fixnum is apparently 2**30 - 1 (or 2**62 - 1), because one bit is used to mark it as an integer instead of an object reference.

Answer 3

Reading the friendly manual? Who'd want to do that?

start = Time.now
largest_known_fixnum = 1
smallest_known_bignum = nil
until (smallest_known_bignum and smallest_known_bignum == largest_known_fixnum + 1)
  if smallest_known_bignum.nil?
    next_number_to_try = largest_known_fixnum * 1000
  else
    next_number_to_try = (smallest_known_bignum + largest_known_fixnum) / 2 #Geometric mean would be more efficient, but more risky
  end
  raise "Can't happen case" if next_number_to_try <= largest_known_fixnum or (smallest_known_bignum and next_number_to_try >= smallest_known_bignum)
  if next_number_to_try.class == Bignum
    smallest_known_bignum = next_number_to_try
  elsif next_number_to_try.class == Fixnum
    largest_known_fixnum = next_number_to_try
  else
    raise "Can't happen case"
  end
end
finish = Time.now
puts "The largest fixnum is #{largest_known_fixnum}"
puts "The smallest bignum is #{smallest_known_bignum}"
puts "Calculation took #{finish - start} seconds"

Answer 4

FIXNUM_MAX = (2**(0.size * 8 -2) -1)
FIXNUM_MIN = -(2**(0.size * 8 -2))