How to work with leading zeros in integers

Question

What is the proper way to deal with leading zeros in Ruby?

0112.to_s
 => \"74\"
0112.to_i
 => 74

Why is it converting 0112

Answer 1

You can't, because Ruby's Integer class does not store leading zeros.

A leading 0 in a number literal is interpreted as a prefix:

• 0 and 0o: octal number
• 0x: hexadecimal number
• 0b: binary number
• 0d: decimal number

It allows you to enter numbers in these bases. Ruby's parser converts the literals to the appropriate Integer instances. The prefix or leading zeros are discarded.

Another example is %w for entering arrays:

ary = %w(foo bar baz)
#=> ["foo", "bar", "baz"]

There's no way to get that %w from ary. The parser turns the literal into an array instance, so the script never sees the literal.

0112 (or 0o112) is interpreted (by the parser) as the octal number 112 and turned into the integer 74.

A decimal 0112 is just 112, no matter how many zeros you put in front:

0d0112   #=> 112
0d00112  #=> 112
0d000112 #=> 112

It's like additional trailing zeros for floats:

1.0   #=> 1.0
1.00  #=> 1.0
1.000 #=> 1.0

You probably have to use a string, i.e. "0112"

Another option is to provide the (minimum) width explicitly, e.g.:

def descending_order(number, width = 0)
  sprintf('%0*d', width, number).reverse.to_i
end

descending_order(123, 4)
#=> 3210

descending_order(123, 10)
#=> 3210000000

Answer 2

A numeric literal that starts with 0 is an octal representation, except the literals that start with 0x which represent hexadecimal numbers or 0b which represent binary numbers.

1 * 8**2 + 1 * 8**1 + 2 * 8**0 == 74

To convert it to 0112, use String#% or Kernel#sprintf with an appropriate format string:

'0%o' % 0112  # 0: leading zero, %o: represent as an octal
# => "0112"