Way to count number of digits in a Bignum

Question

I am writing a ruby method as follows:

def build_array(word)
  word_copy = word
  array = []
  length = word_copy.length
  for i in 1..length do
    array[i] = word_copy % 10
    word_copy = word_copy/10;
  end
 puts array
end

I would like to create an iterator which counts from 1 to the number of digits in a Bignum.

Bignum.length is not valid ruby. Alternatively, is there a way to bust up a Bignum into an array of its constituent digits (Which is what I am trying to implement).

Thanks!

Are you looking for `num.to_s`? – Sergio Tulentsev Dec 21 '15 at 11:31 — Sergio Tulentsev, Dec 21 '15 at 11:31

sawa · Answer 1 · 2015-12-21T12:09:30.653

3

x = 424723894070784320891478329472334238899
Math.log10(x).floor + 1 # => 39

or

Math.log10(x + 1).ceil # => 39

edited Dec 21 '15 at 12:09

answered Dec 21 '15 at 11:38

sawa

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And it's faster too! Although, way less obvious, if I may. – Sergio Tulentsev Dec 21 '15 at 11:48
@SergioTulentsev I thought this was the most straightforward. I cannot think of a more obvious answer than this. – sawa Dec 21 '15 at 11:49
maybe it's just me being bad at maths. – Sergio Tulentsev Dec 21 '15 at 11:49

score 2 · Accepted Answer · answered Dec 21 '15 at 11:32

2

You could convert it to String and count its size

b = 2_999_999_000_000_000
p b.class
#=> Bignum
p b.to_s.size
#=> 16
p b.to_s.split("").map(&:to_i)
#=> [2, 9, 9, 9, 9, 9, 9, 0, 0, 0, 0, 0, 0, 0, 0, 0]

answered Dec 21 '15 at 11:32

Wand Maker

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`b.to_s.chars.map(&:to_i)` saves a few strokes and reduces line noise. – the Tin Man Dec 21 '15 at 22:08

Way to count number of digits in a Bignum

2 Answers2