Without Converting to a String, How Many Digits Does a Fixnum Have?

Question

I want find the length of a Fixnum, num, without converting it into a String.

In other words, how many digits are in num without calling the .to_s() method:

num.to_s.length

Answer 1

puts Math.log10(1234).to_i + 1 # => 4

You could add it to Fixnum like this:

class Fixnum
  def num_digits
    Math.log10(self).to_i + 1
  end
end

puts 1234.num_digits # => 4

Answer 2

Ruby 2.4 has an Integer#digits method, which return an Array containing the digits.

num = 123456
num.digits
# => [6, 5, 4, 3, 2, 1] 
num.digits.count
# => 6 

EDIT:

To handle negative numbers (thanks @MatzFan), use the absolute value. Integer#abs

-123456.abs.digits
# => [6, 5, 4, 3, 2, 1]

Answer 3

Sidenote for Ruby 2.4+

I ran some benchmarks on the different solutions, and Math.log10(x).to_i + 1 is actually a lot faster than x.to_s.length. The comment from @Wayne Conrad is out of date. The new solution with digits.count is trailing far behind, especially with larger numbers:

with_10_digits = 2_040_240_420

print Benchmark.measure { 1_000_000.times { Math.log10(with_10_digits).to_i + 1 } }
# => 0.100000   0.000000   0.100000 (  0.109846)
print Benchmark.measure { 1_000_000.times { with_10_digits.to_s.length } }
# => 0.360000   0.000000   0.360000 (  0.362604)
print Benchmark.measure { 1_000_000.times { with_10_digits.digits.count } }
# => 0.690000   0.020000   0.710000 (  0.717554)

with_42_digits = 750_325_442_042_020_572_057_420_745_037_450_237_570_322

print Benchmark.measure { 1_000_000.times { Math.log10(with_42_digits).to_i + 1 } }
# => 0.140000   0.000000   0.140000 (  0.142757)
print Benchmark.measure { 1_000_000.times { with_42_digits.to_s.length } }
# => 1.180000   0.000000   1.180000 (  1.186603)
print Benchmark.measure { 1_000_000.times { with_42_digits.digits.count } }
# => 8.480000   0.040000   8.520000 (  8.577174)

Answer 4

Although the top-voted loop is nice, it isn't very Ruby and will be slow for large numbers, the .to_s is a built-in function and therefore will be much faster. ALMOST universally built-in functions will be far faster than constructed loops or iterators.

Answer 5

Another way:

def ndigits(n)
  n=n.abs
  (1..1.0/0).each { |i| return i if (n /= 10).zero? }
end

ndigits(1234) # => 4
ndigits(0)    # => 1
ndigits(-123) # => 3

Answer 6

If you don't want to use regex, you can use this method:

def self.is_number(string_to_test)
is_number = false
# use to_f to handle float value and to_i for int
string_to_compare = string_to_test.to_i.to_s
string_to_compare_handle_end = string_to_test.to_i

# string has to be the same
if(string_to_compare == string_to_test)
  is_number = true
end
# length for fixnum in ruby
size = Math.log10(string_to_compare_handle_end).to_i + 1
# size has to be the same
if(size != string_to_test.length)
  is_number = false
end
is_number
end

Answer 7

You don't have to get fancy, you could do as simple as this.

def l(input)
  output = 1
  while input - (10**output) > 0
    output += 1
  end
  return output
end
puts l(456)

Answer 8

It can be a solution to find out the length/count/size of a fixnum.

irb(main):004:0> x = 2021
=> 2021
irb(main):005:0> puts x.to_s.length
4
=> nil
irb(main):006:0>