让我们改进堆栈器的示例，并使用Scala的case类：
case class Person(firstName: String, lastName: String)
上面的Scala类包含下面的Java类的所有功能， 以及更多其他功能 -例如，它支持模式匹配（Java没有此功能）。Scala
2.8添加了命名和默认参数，这些参数用于为案例类生成一个复制方法，它具有与以下Java类的with *方法相同的功能。
public class Person implements Serializable { private final String firstName; private final String lastName; public Person(String firstName, String lastName) { this.firstName = firstName; this.lastName = lastName; } public String getFirstName() { return firstName; } public String getLastName() { return lastName; } public Person withFirstName(String firstName) { return new Person(firstName, lastName); } public Person withLastName(String lastName) { return new Person(firstName, lastName); } public boolean equals(Object o) { if (this == o) { return true; } if (o == null || getClass() != o.getClass()) { return false; } Person person = (Person) o; if (firstName != null ? !firstName.equals(person.firstName) : person.firstName != null) { return false; } if (lastName != null ? !lastName.equals(person.lastName) : person.lastName != null) { return false; } return true; } public int hashCode() { int result = firstName != null ? firstName.hashCode() : 0; result = 31 * result + (lastName != null ? lastName.hashCode() : 0); return result; } public String toString() { return "Person(" + firstName + "," + lastName + ")"; }}然后，在使用中（当然）：
Person mr = new Person("Bob", "Dobbelina");Person miss = new Person("Roberta", "MacSweeney");Person mrs = miss.withLastName(mr.getLastName());反对
val mr = Person("Bob", "Dobbelina")val miss = Person("Roberta", "MacSweeney")val mrs = miss copy (lastName = mr.lastName)
