comparable

问题 Let's say that I have an array with objects, where I have some employees (objects). They all have: int age , double salary . I want to sort this array so my class implements Comparable <Employee> . I've made a method: public int compareTo(Employee other) { return Double.compare(salary, other.salary); } And it's ok, sorting is working fine. But I'm sorting by double salary . Now I want to sort by int age so what now? I've made a method: public int compareAge(Employee other) { return Integer

问题 This question already has answers here : Closed 6 years ago . Possible Duplicate: Java: SortedMap, TreeMap, Comparable? How to use? I am using the Java JungI graph package and Netbeans 7. I am getting the following error from Java: Exception in thread "main" java.lang.ClassCastException: graphvisualization.MyVertex cannot be cast to java.lang.Comparable at java.util.TreeMap.put(TreeMap.java:542) Here is the code associated with the error: SortedMap<MyVertex, Double> vMap = new TreeMap

问题 I want to write a generic Pair class, which has two members: key and value. The only requirement to this class is that both key and value should implements the Comparable interface, otherwise Pair class will not accept them as type parameter. First I code it like this: public class Pair<T1 extends Comparable, T2 extends Comparable> But the JDK 1.6 compiler will generate warning about this: Comparable is a raw type. References to generic type Comparable<T> should be parameterized Then I tried

问题 I was wondering why the sort method of the Arrays class is asking for a parameter of type Object[]. Why the parameter is not of type Comparable[]. If you don't pass a Comparable[] it's generating a ClassCastException. Why ... public static void sort(Object[] a) and not public static void sort(Comparable[] a) ? Thanks 回答1: Because the second form would require a reallocation of the array. Even if you know that your array contains only comparables, you cannot just cast it to Comparable[] if the

问题 I need an example on how to use a comparable class on a HashSet to get an ascending order. Let’s say I have a HashSet like this one: HashSet<String> hs = new HashSet<String>(); How can I get hs to be in ascending order? 回答1: Use a TreeSet instead. It has a constructor taking a Comparator. It will automatically sort the Set . If you want to convert a HashSet to a TreeSet , then do so: Set<YourObject> hashSet = getItSomehow(); Set<YourObject> treeSet = new TreeSet<YourObject>(new YourComparator

问题 An enum in Java implements the Comparable interface. It would have been nice to override Comparable 's compareTo method, but here it's marked as final. The default natural order on Enum 's compareTo is the listed order. Does anyone know why a Java enums have this restriction? 回答1: For consistency I guess... when you see an enum type, you know for a fact that its natural ordering is the order in which the constants are declared. To workaround this, you can easily create your own Comparator

问题 Comparable contract specifies that e.compareTo(null) must throw NullPointerException . From the API: Note that null is not an instance of any class, and e.compareTo(null) should throw a NullPointerException even though e.equals(null) returns false . On the other hand, Comparator API mentions nothing about what needs to happen when comparing null . Consider the following attempt of a generic method that takes a Comparable , and return a Comparator for it that puts null as the minimum element.

问题 I have a class public class StudentVO { int age; String name; } I used the same class in two different areas. At one place i need to sort based on the age. In another place I need to sort based on the name and in another place i may need sorting based on both age and name. How can I do this? If one field I can override compareTo() . Is it possible to do this? 回答1: 1)You should write two Comparator for sorting on age and name separately, and then use the Collections.sort(List,Comparator).

问题 This question already has answers here : Comparing the values of two generic Numbers (12 answers) Closed 2 years ago . Does anyone know why java.lang.Number does not implement Comparable ? This means that you cannot sort Number s with Collections.sort which seems to me a little strange. Post discussion update: Thanks for all the helpful responses. I ended up doing some more research about this topic. The simplest explanation for why java.lang.Number does not implement Comparable is rooted in

问题 This question already has answers here : Comparing the values of two generic Numbers (12 answers) Closed 2 years ago . Does anyone know why java.lang.Number does not implement Comparable ? This means that you cannot sort Number s with Collections.sort which seems to me a little strange. Post discussion update: Thanks for all the helpful responses. I ended up doing some more research about this topic. The simplest explanation for why java.lang.Number does not implement Comparable is rooted in