Sign Up

Sign Up to our social questions and Answers Engine to ask questions, answer people’s questions, and connect with other people.

Have an account? Sign In

Have an account? Sign In Now

Sign In

Login to our social questions & Answers Engine to ask questions answer people’s questions & connect with other people.

Sign Up Here

Forgot Password?

Don't have account, Sign Up Here

Forgot Password

Lost your password? Please enter your email address. You will receive a link and will create a new password via email.

Have an account? Sign In Now

You must login to ask a question.

Forgot Password?

Need An Account, Sign Up Here

Please briefly explain why you feel this question should be reported.

Please briefly explain why you feel this answer should be reported.

Please briefly explain why you feel this user should be reported.

Sign InSign Up

The Archive Base

The Archive Base Logo The Archive Base Logo

The Archive Base Navigation

  • SEARCH
  • Home
  • About Us
  • Blog
  • Contact Us
Search
Ask A Question

Mobile menu

Close
Ask a Question
  • Home
  • Add group
  • Groups page
  • Feed
  • User Profile
  • Communities
  • Questions
    • New Questions
    • Trending Questions
    • Must read Questions
    • Hot Questions
  • Polls
  • Tags
  • Badges
  • Buy Points
  • Users
  • Help
  • Buy Theme
  • SEARCH
Home/ Questions/Q 546443
In Process

The Archive Base Latest Questions

Editorial Team
  • 0
Editorial Team
Asked: May 13, 20262026-05-13T10:52:03+00:00 2026-05-13T10:52:03+00:00

Is there a simple way to return regex matches as an array? Here is

  • 0

Is there a simple way to return regex matches as an array?
Here is how I am trying in 2.7.7:

val s = """6 1 2"""
val re = """(\d+)\s(\d+)\s(\d+)""".r
for (m <- re.findAllIn (s)) println (m) // prints "6 1 2"
re.findAllIn (s).toList.length // 3? No! It returns 1!

But I then tried:

s match {
  case re (m1, m2, m3) => println (m1)
}

And this works fine! m1 is 6, m2 is 1, etc.

Then I found something that added to my confusion:

val mit = re.findAllIn (s)
println (mit.toString)
println (mit.length)
println (mit.toString)

That prints:

non-empty iterator
1
empty iterator

The “length” call somehow modifies the state of the iterator. What is going on here?

  • 1 1 Answer
  • 0 Views
  • 0 Followers
  • 0
Share
  • Facebook
  • Report

Leave an answer
Cancel reply

You must login to add an answer.

Forgot Password?

Need An Account, Sign Up Here

1 Answer

  • Voted
  • Oldest
  • Recent
  • Random
  1. Editorial Team
    Editorial Team
    2026-05-13T10:52:04+00:00Added an answer on May 13, 2026 at 10:52 am

    Ok, first of all, understand that findAllIn returns an Iterator. An Iterator is a consume-once mutable object. ANYTHING you do to it will change it. Read up on iterators if you are not familiar with them. If you want it to be reusable, then convert the result of findAllIn into a List, and only use that list.

    Now, it seems you want all matching groups, not all matches. The method findAllIn will return all matches of the full regex that can be found on the string. For example:

    scala> val s = """6 1 2, 4 1 3"""
    s: java.lang.String = 6 1 2, 4 1 3
    
    scala> val re = """(\d+)\s(\d+)\s(\d+)""".r
    re: scala.util.matching.Regex = (\d+)\s(\d+)\s(\d+)
    
    scala> for(m <- re.findAllIn(s)) println(m)
    6 1 2
    4 1 3
    

    See that there are two matches, and neither of them include the “, ” at the middle of the string, since that’s not part of any match.

    If you want the groups, you can get them like this:

    scala> val s = """6 1 2"""
    s: java.lang.String = 6 1 2
    
    scala> re.findFirstMatchIn(s)
    res4: Option[scala.util.matching.Regex.Match] = Some(6 1 2)
    
    scala> res4.get.subgroups
    res5: List[String] = List(6, 1, 2)
    

    Or, using findAllIn, like this:

    scala> val s = """6 1 2"""
    s: java.lang.String = 6 1 2
    
    scala> for(m <- re.findAllIn(s).matchData; e <- m.subgroups) println(e)
    6
    1
    2
    

    The matchData method will make an Iterator that returns Match instead of String.

    • 0
    • Reply
    • Share
      Share
      • Share on Facebook
      • Share on Twitter
      • Share on LinkedIn
      • Share on WhatsApp
      • Report

Sidebar

Related Questions

I'm wondering what's the best way -- or if there's a simple way with
I'm trying to understand regex as much as I can, so I came up
yes it's another .net regex question :) (please excuse the long waffle leading up
I need to extract the name of the direct sub directory from a full
Using the following text as a sample, I need to be able to extract
I have a Django model with a large number of fields and 20000+ table
I'm using the Jison parser generator for Javascript and am having problems with my
I need to compare 2 strings as equal such as these: Lubeck == Lübeck
mmmh guys, i really hope my english is good enaught to explain what i

Explore

  • Home
  • Add group
  • Groups page
  • Communities
  • Questions
    • New Questions
    • Trending Questions
    • Must read Questions
    • Hot Questions
  • Polls
  • Tags
  • Badges
  • Users
  • Help
  • SEARCH

Footer

© 2021 The Archive Base. All Rights Reserved
With Love by The Archive Base

Insert/edit link

Enter the destination URL

Or link to existing content

    No search term specified. Showing recent items. Search or use up and down arrow keys to select an item.