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    <p>The quick answer is that you call the constructor with your parameters. In Java: </p> <pre><code>public class Contact { private final String firstName; private final String middleName; private final String lastName; private final Date birthday; public Contact( String firstName, String middleName, String lastName, Date birthday ) { this.firstName = firstName; this.middleName = middleName; this.lastName = lastName; this.birthday = birthday; } … // getters } … inside while loop: contacts.add(new Contact( rs.getString("first_name"), rs.getString("middle_name"), rs.getString("last_name"), new Date(birthday.getTime()) ); </code></pre> <p>I left out the optional birthday for now, we'll add it back in the Scala version.</p> <p>We still have a mutable Contacts list though, so lets look at making it properly immutable. We'll use a Scala list. First, the scala data object:</p> <pre><code>case class Contact( first: String, middle: String, last: String, birthday: Option[Date]) </code></pre> <p>We have optional birthday as part of the type-sig as a bonus.</p> <p>Now lets define a simple extractor method, given a resultSet:</p> <pre><code>def contacts(rs: ResultSet) = { @annotation.tailrec def loop(cs: List[Contact]): List[Contact] = if (!rs.next()) cs else loop( Contact( rs.getString("first_name"), rs.getString("middle_name"), rs.getString("last_name"), Option(rs.getLong("birthday")) ) :: cs) loop(Nil) } </code></pre> <p>This version recursively builds up a singly-linked list of all contacts in your ResultSet. This is immutable.</p>
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    1. POHow can I replace a java mutable class with its immutable couterpart in scala?
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    1. USJed Wesley-Smith
    1. USJed Wesley-Smith
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