equals

问题 I have Hashmap,how to override equals method for the hashmap? Thanks. 回答1: if you want you can do: HashMap<String, String> map = new HashMap<String, String>(){ @Override public boolean equals(Object o) { // TODO comparison here return super.equals(o); } }; map.equals(new HashMap<String, String>()); 回答2: In HashMap already equals() method is overrriden.If you want to override object class equals() method eclipse shortcut key--->alt+shift+s+v Select the equals method click on ok. 回答3: You will

问题 I'm trying to filter the user's input to make sure its a max of 4 digits and it's not an empty string. Whether I leave it blank or I enter a number, !strInput.equals(null) still comes up true. Am I comparing the string incorrectly? I also tried: !strInput.equals("") , strInput != null , and strInput != "" though I think it should be .equals(...) since I'm trying to compare values. private void updateFast() { String strInput = JOptionPane.showInputDialog(null, "How many?"); if (!strInput

问题 Why do the equals methods implemented by Eclipse compare each value, wouldn't it be simpler to just compare the hashCodes of both objects? From what I know: hashCode always generates the same hash for the same input So if two objects are equal, they should have the same hash If objects that are equal have the same hash, I can just check the hash in order to determine of objects are equal or not edit: Related question, why does one always implement the hashCode when equals is implemented, if

问题 I have this string from mysql DB: it should be this: 2100428169/2010 this is my code String str = rs.getString("str"); str = str.replaceAll("\\s+",""); str = str.trim(); char[] strCH = str.toCharArray(); and I get this: [, 2, 1, 0, 0, 4, 2, 8, 1, 6, 9, /, 2, 0, 1, 0] Why? It's a problem because I need to use str1.equals(str) but it doesn't work because after Object obj = (object)str; It is in obj again with a space at the beginning like when I use toCharArray so it means equals doesn't work.

问题 I have been working on a vbs login and signup program I have a fully functional working signup program but I am having troubles seeing if a inputbox a.k.a the password is equal to the logged password from signup in a file uname = inputbox("Please type your Username.") pword = inputbox("Please ebter your Password.") Set objFileToRead = CreateObject("Scripting.FileSystemObject").OpenTextFile(uname + ".txt",1) strFileText = objFileToRead.ReadLine() objFileToRead.Close Set objFileToRead = Nothing

问题 I am trying to avoid an infinite loop and am having trouble figuring out whats wrong. This is supposed to find a solution for a 3x2 puzzle board. I suspect the problem may be with my overridden equals method but I'm not sure. Running into two issues: 1) It keeps re-exploring already explored nodes. 2) The queue is empty before a solution is found, causing an error. Driver class: import java.util.*; public class Driver { public static void main(String[] args){ Node test = new Node(new int[]{1,

问题 I've came across this piece of code. I have never seen an equals implemented in such a way. What strike me was that it is really "neat", in the sense that is only takes one line of boilerplate. However, the fact I have never seen this approach before make me suspicious. According to the contract of Java equals and hashCode, is the following implementation legal? @Override public boolean equals(Object o) { return this == o || o instanceof DetailsPageTrackingRequest && this.hashCode() == o

问题 This question already has an answer here : Should Equality Comparison of Float / Double Instance Variables in an Equals Method be Exact? (1 answer) Closed 5 months ago . An object of class Foo is considered equal if the double members are within a given range of the other object. Such an error can easily be introduced due to floating point arithmetic. The method isDoubleEquals and doubleArrayEquals will take care of the equals part but the contract states that the hashcode has to be identical

问题 This question already has an answer here : Should Equality Comparison of Float / Double Instance Variables in an Equals Method be Exact? (1 answer) Closed 5 months ago . An object of class Foo is considered equal if the double members are within a given range of the other object. Such an error can easily be introduced due to floating point arithmetic. The method isDoubleEquals and doubleArrayEquals will take care of the equals part but the contract states that the hashcode has to be identical

问题 Here are my test for two equals methods: Random generator = new Random(); long startTime = System.nanoTime(); for(int i = 0; i<1000; i++) { Integer int11 = generator.nextInt(1000); Integer int22 = generator.nextInt(1000); int11.equals(int22); } long endTime = System.nanoTime(); long duration = (endTime - startTime); System.out.println(duration + " ns"); Random generator1 = new Random(); long startTime1 = System.nanoTime(); for(int i = 0; i<1000; i++) { Integer int1 = generator1.nextInt(1000);