equals

what is the difference between == operator and equals()? (with hashcode() ???)

假装没事ソ 提交于 2019-11-26 23:08:31
I was learning hashcode in more depth and figured that: 1. If you override equals(), you must override hashcode() too. 2. To find if 2 objects are same object, use == operator Given those 2 factors, in Java I was assuming that when == operator is used to compare if 2 instances are same or not, if(object1 == object2) is actually doing if(object1.hashcode() == object2.hashcode()) But it appears I was wrong by running the test below. public class Main { public static void main(String[] args){ Obj1 one = new Obj1(); Obj1 two = new Obj1(); //is this calling hashCode() in backend??? if(one == two) {

Force a class to override the .equals method

…衆ロ難τιáo~ 提交于 2019-11-26 22:42:46
问题 I have a bunch of class who implement a common interface : Command. And this bunch of class goes to a Map. To get the Map working correctly, I need to each class who implements Command to override the Object.equals(Object other) method. it's fine. But i whould like to force the overriding of equals. => Have a compilation error when something who implement command dont override equals. It's that possible ? Edit : BTW , i will also need to forcing the override of hashcode... 回答1: No, you can't.

String.Equals() not working as intended

偶尔善良 提交于 2019-11-26 22:38:32
I am using LINQ to search through one of my Entity Framework tables and find a "group" based on the name. The name is a string and appears to be Unicode (says it is in the edmx). I have a method GetGroup() and I pass in a name to search for. Debugging through the code, I already have a group named "Test" in my database. Once I pass in a group named "TEST" I expect it to return the "Test" which was already in the database. It for some reason, does not find the "Test" and thinks "TEST" doesn't exist. Here is my query, I cannot see why it does not work. Please help. "name" is the passed in the

Operator Overloading with Interface-Based Programming in C#

只愿长相守 提交于 2019-11-26 22:37:02
问题 Background I am using interface-based programming on a current project and have run into a problem when overloading operators (specifically the Equality and Inequality operators). Assumptions I'm using C# 3.0, .NET 3.5 and Visual Studio 2008 UPDATE - The Following Assumption was False! Requiring all comparisons to use Equals rather than operator== is not a viable solution, especially when passing your types to libraries (such as Collections). The reason I was concerned about requiring Equals

Why should I override hashCode() when I override equals() method?

大城市里の小女人 提交于 2019-11-26 22:32:54
Ok, I have heard from many places and sources that whenever I override the equals() method, I need to override the hashCode() method as well. But consider the following piece of code package test; public class MyCustomObject { int intVal1; int intVal2; public MyCustomObject(int val1, int val2){ intVal1 = val1; intVal2 = val2; } public boolean equals(Object obj){ return (((MyCustomObject)obj).intVal1 == this.intVal1) && (((MyCustomObject)obj).intVal2 == this.intVal2); } public static void main(String a[]){ MyCustomObject m1 = new MyCustomObject(3,5); MyCustomObject m2 = new MyCustomObject(3,5);

Overloading operator== versus Equals()

陌路散爱 提交于 2019-11-26 21:53:57
I'm working on a C# project on which, until now, I've used immutable objects and factories to ensure that objects of type Foo can always be compared for equality with == . Foo objects can't be changed once created, and the factory always returns the same object for a given set of arguments. This works great, and throughout the code base we assume that == always works for checking equality. Now I need to add some functionality that introduces an edge case for which this won't always work. The easiest thing to do is to overload operator == for that type, so that none of the other code in the

why we need to override equals and hashcode in java and why cannot we use Object class implementation

試著忘記壹切 提交于 2019-11-26 21:46:22
问题 guys please let me know, in real world why we need to override equals and hashcode and cant we use Object's equals and hashcode. 回答1: Object's equals/hashcode implementation is fine - if you want "reference identity" as your equality. In other words, on object will always compare as equal to itself, but different to another object. If, however, you want two distinct objects to be equal, you've got to override the method to say how they should be equal (and then override hashcode to be

Setting equal heights for div's with jQuery

佐手、 提交于 2019-11-26 21:42:31
I want to set equal height for divs with jQuery. All the divs may have different amount of content and different default height. Here is a sample of my html layout: <div class="container"> <div class="column">This is<br />the highest<br />column</div> <div class="column">One line</div> <div class="column">Two<br />lines</div> </div> <div class="container"> <div class="column">One line</div> <div class="column">Two<br>lines</div> <div class="column">One line</div> </div> I'm setting the height with the next jQuery function: $(document).ready(function(){ var highestBox = 0; $('.container .column

How equals() method works

妖精的绣舞 提交于 2019-11-26 21:36:02
问题 I am digging into the basics of Java. I infer from this article, that java equals method means, if two objects are equal then they must have the same hashCode(). Here's my example. public class Equals { /** * @param args */ public static void main(String[] args) { String a = new String("a"); String b = new String("a"); System.out.println("a.hashCode() "+a.hashCode()); System.out.println("b.hashCode() "+b.hashCode()); System.out.println(a == b); System.out.println(a.equals(b)); } } output: a

Why equals and hashCode were defined in Object?

孤街醉人 提交于 2019-11-26 21:14:01
问题 What's the reasoning behind decision to include these methods in the java.lang.Object? Equality and hashing doesn't make sense for many classes. It would be more logical to make two interfaces: interface Equalable { boolean equals(Equalable other); } interface Hashable extends Equalable { int hashCode(); } For example HashSet definition might look like class HashSet<T extends Hashable> ... It would prevent one of the common beginner mistakes - using set of items without implementing equals