How to work with leading zeros in integers

Question

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

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

Why is this converting 0112 to 74 ?

How to convert 0112 to the string "0112" ?

---

I want to define a method that takes an integer as an argument and returns it with its digits in descending order.

But this does not work for me when I have leading zeros:

 def descending_order(n) n.to_s.reverse.to_i end 

Answer 1

A numeric literal starting with 0 is an octal representation, with the exception of literals starting 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 the appropriate format string:

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

Answer 2

You cannot, because the Ruby Integer class does not store leading zeros.

The leading 0 in the numeric literal is interpreted as the prefix:

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

It allows you to enter numbers in these databases. The Ruby parser converts literals to their corresponding Integer instances. The prefix or leading zeros are discarded.

Another example is %w for inputting arrays:

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

Unable to get this %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 turns into an integer 74 .

Decimal 0112 is just 112 , no matter how many zeros you set in front of you:

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

It looks like extra trailing zeros for floats:

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

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

Another option is to explicitly specify the minimum width, for example:

 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