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Cannot Be Cast To Java.lang.comparable Treeset


Book Review: Murach's Java Servlets and JSP Phobos - A JavaFX Games Engine: Part 2 - JavaFX Scene API and the FSM Maven Tutorial 2 - Adding Dependencies Maven Tutorial 1 new TreeMap(new Comparator() { public int compare(MyVertex o1, MyVertex o2) { } }); share|improve this answer answered Jan 3 '13 at 6:33 Kumar Vivek Mitra 25.2k52554 add a comment| Not Browse other questions tagged java or ask your own question. Java Collection classes like HashMap, ArrayList, Vector or Hashtable which is not using Generics. 2. http://haywirerobotics.com/cannot-be/operator-cannot-be-applied-to-java-lang-string.html

Generics are designed to write type-safe code and provides compile time checks which tends to violate type-safety. I couldn't tell you for sure with you code though as it's incomplete. Other Java debugging and troubleshooting tutorials How to solve java.util.NoSuchElementException in Java JDBC - java.lang.ClassNotFoundException: com.mysql.jdbc.Driver Solution How to fix java.io.NotSerializableException: org.apache.log4j.Logger Error in Java How to fix java.sql.SQLException: Invalid column Reference Sheets Code Snippets C Snippets C++ Snippets Java Snippets Visual Basic Snippets C# Snippets VB.NET Snippets ASP.NET Snippets PHP Snippets Python Snippets Ruby Snippets ColdFusion Snippets SQL Snippets Assembly Snippets

Cannot Be Cast To Java.lang.comparable Treeset

I think that's what's tripping you up. Notice how the cast gets moved to somewhere you did not explicitly put it. You killed my father. super T" is to ensure that T can also be java.sql.Timestamp which is a Comparable.

share|improve this answer answered Jan 21 '10 at 20:11 Powerlord 60.7k1193149 1 "I took a look at some of my classmates' source codes" I'm thinking that this class is his share|improve this answer edited Jan 21 '10 at 21:10 answered Jan 21 '10 at 20:32 Paul Wagland 14.1k73759 add a comment| up vote 0 down vote I haven't put too much What is this? Java Comparable Or is it the ?

a) The bag is an "array of Object". Ljava Lang Object Cannot Be Cast To Ljava Lang Comparable Check out the Collections class.... posted 6 years ago Rob Prime wrote:That said, the safest choice for the created array type is what T extends. http://stackoverflow.com/questions/27591061/ljava-lang-object-cannot-be-cast-to-ljava-lang-integer Is Area of a circle always irrational This is my pillow How to decline a postdoc interview if there is some possible future collaboration?

Maybe it's nitpicking, but I don't agree with the underlined reasoning: you cannot "add anything", as the cast will fail. Compareto Java After all, you know that every T is a Comparable. The declaration should be public class MyArrayList

Ljava Lang Object Cannot Be Cast To Ljava Lang Comparable

When you write Stack st = new Stack();, java understands it as Stack st = new Stack();, it causes your Ljava.lang.Object; cannot be cast to [Ljava.lang.Integer error. The declaration of E is protected E[] elements = null; This is how I am trying to call Random ran = new Random(); Stack st = new Stack(); st.push(ran.nextInt(100)); Update Guys, Cannot Be Cast To Java.lang.comparable Treeset Let me be more specific: what I don't understand is the line: Comparable temp = (Comparable) list[loc] It looks as if we are casting the object located at list[loc] to an Create Generic Array Java How to compare Arrays in Java - Equals vs deepEqua...

I'm not sure why it's "extends Comparable" though I expect it's for a similar reason that interfaces do not implement each other, but rather extend each other. http://haywirerobotics.com/cannot-be/the-type-java-lang-object-cannot-be-resolved-it-is-indirectly-referenced-from-class-files.html Comparable is an interface, is it not? Draw a hollow square of # with given width Why are password boxes always blanked out when other sensitive data isn't? A problem only occurs when the (incorrect) fact that elements is type E[] is "exposed" to the outside of the class, outside of the scope of the erasure of E, into Treeset Comparator

If so, how can you then successfully compare this object that was cast to Comparable if the method compareTo() is not implemented (interfaces are classes with no implemented methods, correct?)? asked 3 years ago viewed 9921 times active 4 months ago Upcoming Events 2016 Community Moderator Election ends Nov 22 Linked 4 Java: SortedMap, TreeMap, Comparable? Why there are no approximation algorithms for SAT and other decision problems? http://haywirerobotics.com/cannot-be/dependencies-cannot-be-applied-to-groovy-lang-closure-java.html That should solve the warning, but the ClassCastException will still exist.

It's an Object[] and the cast will fail. Java Queue So we first create a reference of the type Comparable and then use the compareTo() method. Seems pretty straightforward, but I've been working on it for hours now.

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extends T> coll) For a thorough explanation, I recommend Java Generics and Collections by Maurice Naftalin Maurice and Philip Wadler. You're in a darkened room. Thanks for any tips on this. If you want to sort using different order then you need to provide your own comparator.

import java.util.TreeSet; class Dog { int size; Dog(int s) { size = s; } } public class ImpComparableWrong { public static void main(String[] args) { TreeSet i = Now I tell you, "the bag is full of peanut M&Ms." You say "cool!," grab a handful, and toss them into your mouth. Mapping network drive in Windows XP and 7 - net us... have a peek here I have looked and can not figure it out or find information on it that is helpful.

Or is it the ? Or you could set the report to use Groovy instead of Java for expressions. Implement Comparable interface. if(temp.compareTo(searchItem) >= 0) { found = true; break; } } if(found) { if(list[loc].equals(searchItem)) return loc; else return -1; } else return -1; } Hello.