How do I parse a number from a String that may have a leading zero?

Question

In ruby I am parsing a date in the following format: 24092008. I want to convert each section (year, month, date) into a number.

I have split them up using a regex which produces three Strings which I am passing into the Integer constructor.

  date =~ /^([\d]{2})([\d]{2})([\d]{4})/
  year = Integer($3)
  month = Integer($2)
  day = Integer($1)

When it hits the month line it crashes as follows:

`Integer': invalid value for Integer: "09" (ArgumentError)

It took me a while to realise that it's interpreting the leading zero as Octal and 09 is not a valid Octal number (it works fine with "07").

Is there an elegant solution to this or should I just test for numbers less than 10 and remove the zero first?

Thanks.

Answer 1

I'm not familiar with regexes, so forgive me if this answer's off-base. I've been assuming that $3, $2, and $1 are strings. Here's what I did in IRB to replicate the problem:

irb(main):003:0> Integer("04")
=> 4
irb(main):004:0> Integer("09")
ArgumentError: invalid value for Integer: "09"
    from (irb):4:in `Integer'
    from (irb):4
    from :0

But it looks like .to_i doesn't have the same issues:

irb(main):005:0> "04".to_i
=> 4
irb(main):006:0> "09".to_i
=> 9

Answer 2

Specify base 10

Tell Ruby explicitly that you want to interpret the string as a base 10 number.

Integer("09", 10) # => 9

This is better than .to_i if you want to be strict.

"123abc".to_i # => 123
Integer("123abc", 10) # => ArgumentError

How I figured this out

In irb, method(:Integer) returns #<Method: Object(Kernel)#Integer>. That told me that Kernel owns this method, and I looked up the documentation on Kernel. The method signature shows that it takes a base as the second argument.

Answer 3

Perhaps (0([\d])|([1-9][\d])) in place of ([\d]{2}) You may have to use $2, $4, and $5 in place of $1, $2, $3.

Or if your regexp supports (?:...) then use (?:0([\d])|([1-9][\d]))

Since ruby takes its regexp from perl, this latter version should work.