This question already has answers here:
How do I compare strings in Java?
(23 answers)
Closed 5 years ago.
public void play () {
int anInteger;
//guess return code
int code;
while (true) {
String input=null;
input = JOptionPane.showInputDialog("Please enter an integer");
if (input == "-1") {
//JOptionPane.showMessageDialog(null, input);
System.exit(0);
break;
} else {
if (input==null) {
System.exit(0);
} else if (input.isEmpty()) {
continue;
} else {
anInteger = Integer.parseInt(input);
code = this.oneGuess (anInteger);
//JOptionPane.showMessageDialog(null, anInteger);
}
}
}
}
I want, if the user enter -1, show the program will not prompt the message box any more. Above is the code I have come up with, so far. Why it doesn't work?
String comparisons does NOT work with "==" operator, use "String.equals(Object)" function
input.equals("-1");
Better way would be
"-1".equals(input);
as it also takes care of null input
You are comparing strings, which are objects, with the == operator, which checks whether two object references refer to the same object instance. Instead you should use the equals method for comparing them.
There is a difference between comparing with == and equals. The first compares pointers, the latter contents. That is probably your issue.
You compare Strings with ==, which creates a problem. You can have many different String-Objects which all show "-1". The == tests, if you have exactly the same object on the left and right side. You want to know, if the objects on the left and right sie have an equal content.
Better try
input.equalsIgnoreCase("-1");
EDIT: To answer the comment: input.equalsIgnoreCase("-1") is the same as input.equals("-1") in the case of "-1" as there are no uppercase/lowercase letters in "-1". However, I prefer equalsIgnoreCase in the case of Strings, because it is defined on String, rather than on Object. Still, as the equals-definition is overridden for the String class, it works too in this example and "ignoreCase" is not needed.
Related
This question already has answers here:
What is the difference between == and equals() in Java?
(26 answers)
Closed 10 months ago.
I am within my first year of CS and near the end of my first Java themed course so I'm not quite sure how to find the answer to my question myself.
While writing some code for a project I created my input scanner as:
Scanner scanner = new Scanner(System.in);
I am taking user inputs as strings via a variable assignment:
String userInput = scanner.nextLine();
the user should only be entering strings of char "1" - "6" and "q" (to quit app)
What I'm using that works currently is as follows:
userInput = scanner.nextLine();
while (!appQuit) { //So long as user doesn't quit application
if (userInput.equals("q")) {
appQuit = true;
}
else if (userInput.equals("1")) { //Menu selection for intake a new dog
intakeNewDog(scanner);
displayMenu();
userInput = scanner.nextLine();
}
//removed "2" - "6" for brevity
else {
System.out.println("Not a valid input");
displayMenu();
userInput = scanner.nextLine();
}
}
The only way I found to check equality was the userInput.equals() function.
When I originally wrote it I tried using:
if (userInput == "1") { code }
but it would never successfully compare values as I thought it would.
Any insight into why one method works over the other? Or where I should be looking for these answers?
-Jonesy
The == equal operator compares the object references where the equals function compares the value.
For primitive types and enums the == equal operator compares the value.
An exception happens for comparing strings in a switch case statement since it internal uses the equals method.
As a rule of thumb, always use equals comparison for String. There maybe is, but i have not seen a case where reference comparison was important.
https://docs.oracle.com/javase/8/docs/technotes/guides/language/strings-switch.html
Also interesting:
What makes reference comparison (==) work for some strings in Java?
In simple words, == checks if both objects point to the same memory location whereas . equals() evaluates to the comparison of values in the objects. You can override the equals method to do more specific things, but that's the just of it.
This is literally the first thing that appears if you search java == vs equals in google.
While you might be trying to compare two strings, the operator == does not behave in java as it does in other languages.
This question already has answers here:
How do I compare strings in Java?
(23 answers)
Closed 9 years ago.
So, for some reason I'm having problems just USING a string input.
I don't know why. Maybe it's some incredibly stupid thing everyone knows, but I don't.
Here's the non-functioning code:
import javax.swing.*;
public class Thing {
public static void main(String[] args) {
String input;
JOptionPane.showMessageDialog(null,"Welcome to the test...");
input = JOptionPane.showInputDialog("Do you wish to take the tutorial?" + "\n" +
"If affirmative, enter 'Yes'");
String i = input;
if(i == "Yes") {
tutorial();
} else if(input=="'Yes'") {
JOptionPane.showMessageDialog(null,"Don't actually put apostraphes around you're answer.");
tutorial();
} else {
JOptionPane.showMessageDialog(null,"Remember, you can pull up the tutorial at any time with 'T'");
}
}
Yes, I actually do have a tutorial method somewhere else, and it works fine.
The main problem is that if I enter 'Yes' or Yes, it still goes to the final else.
I only put in the
String i = input;
and changed it from
if(input == "Yes") {
because it didn't work then, either.
So what am I doing wrong?
Don't use the == operator to compare Strings, use equals() instead, as thoroughly explained here, here, here, here or any of the numerous duplicates.
if ("Yes".equals(input))
Or even
if ("yes".equalsIgnoreCase(input))
Notice that the operation is invoked on the "yes" literal to avoid a possible NullPointerException in the case input was null and the operation was invoked on it (Yoda condition).
From the Java Language Specification, Chapter 15 - Expressions, section 21 - Equality Operators:
15.21.3. Reference Equality Operators == and !=
While == may be used to compare references of type String, such an equality test determines whether or not the two operands refer to the same String object. The result is false if the operands are distinct String objects, even if they contain the same sequence of characters (ยง3.10.5). The contents of two strings s and t can be tested for equality by the method invocation s.equals(t).
As mentioned, the problem is that you are comparing this String using the == comparator, not the .equals() method.
If you are running on Java 7, my advice, for a cleaner solution, would be also to wrap this in a switch statement:
JOptionPane.showMessageDialog(null,"Welcome to the test...");
String input = JOptionPane.showInputDialog("Do you wish to take the tutorial?" + "\n" +
"If affirmative, enter 'Yes'");
switch (input) {
case "Yes":
tutorial();
break;
case "'Yes'":
JOptionPane.showMessageDialog(null,"Don't actually put apostraphes around you're answer.");
tutorial();
break;
default:
JOptionPane.showMessageDialog(null,"Remember, you can pull up the tutorial at any time with 'T'");
}
This question already has answers here:
How do I compare strings in Java?
(23 answers)
Closed 9 years ago.
I am still new to Java. I am trying to create a program where users must answer a multiple choice quiz. The users will input their answer and these inputs will be formed into an array. Then I plan on using a for loop to compare the user's answers array to the array of correct answers to tell the user if they are right or wrong.
However it seems that i am not properly comparing the 2 arrays in my if statement. Every time I run the program, it goes straight to the else statements.
My hunch is that the scanner class does not actually store the value?
Can anyone help?
Part of code below:
//Above this section is just a collection of "System.out.println" statements that state questions and answers the user choose from.
int x;
String answers [] = {"a", "a", "b"};
//answers array has the correct answer
Scanner in = new Scanner(System.in);
String answerEntered [] = new String [5];
//user input will be in this arra
for(x=0 ; x<3 ; x++)
{
System.out.print((1+x)+". ");
answerEntered[x] = in.nextLine();
}
for( x=0; x<3; x++)
{
**if(answerEntered[x] == answers[x])
{
System.out.println("For Question "+(x+1)+", you are Correct!");
}**
//This if section does not seem to work. Every time i run the code it automatically goes to the else statement.
else
{
System.out.println("The correct answer for Question "+(x+1)+" is: "+answers[x]);
}
}
In Java, String aren't primitive values, you have to use String.equals() to compare strings
so, change this:
if(answerEntered[x] == answers[x])
to
if(answerEntered[x].equals(answers[x]))
I would also suggest that you check for nullability and ignore case, so:
String answer = answerEntered[x];
boolean isAnswerCorrect =
answer != null &&
//remove trailling spaces and ignore case (a==A)
answer.trim().equalsIgnoreCase(answers[x]);
if(isAnswerCorrect){
For String or any object-equality test in Java, you should almost always be using equals. The == operator only compares references when used with objects (but will work the way you expect it to with primitives like int, boolean, etc); that is, it checks to see if the operands both point/refer to the same object instance. What you're interested in doing is comparing the contents of the String instance, and equals will do that for you:
if(answerEntered[x].equals(answers[x])) {
...
}
For String comparison, you need to use equals instead of ==, which for non-primitive data types, such as String, compares their references, not values.
String a = "foo";
String b = "bar";
if (a.equals(b))
{
//doSomething
}
The problem is in the comparison :
String a = "foo";
String b = "bar";
if (a.equals(b))
//doSomething
AS it Has been answered before.
Extra information, in the for loop of the if / else you are looping only the first 3 positions, not the 5 that exists in the answerEntered array.
Cheers
Use .equals to compare strings. equals compares the values, where == compares the reference
.
In Java, == comparison compares reference identity, means the two things you compare must be the same object. Two objects with the same values are treated as different.
You statement:
if(answerEntered[x] == answers[x])
The answerEntered contains string that is different with any string in answer even if they have the same value.
Java uses Object's .equals method to compare by value, i.e. Two objects are equal as long as they have the same value.
Changing:
if(answerEntered[x] == answers[x])
to
if(answerEntered[x].equals(answers[x]))
should solve the problem.
Also, as answerEntered contains user inputed value, you'd better pre-process it before using it. For example, user might put answer "a " with spaces at the end. You might want to get rid of those spaces as well.
Otherwise "a " will be treated as an incorrect answer.
This question already has answers here:
Closed 10 years ago.
Possible Duplicate:
How do I compare strings in Java?
So, I have a question regarding the testing operators and strings. I'm trying to write a program that will take in user input of either "boy" or "girl". When the user inputs "boy", the output should be "You are a boy." When the user inputs "girl", the output should be "You are a girl."
However, when I compile and run the program, no matter what I input, the output is always "You are a girl."
Is this because the strings have no actual value like integers and therefore, testing operators cannot be used to compare them?
Also, is there anything like the assert function from python in java?
import java.util.Scanner;
class apples{
public static void main(String args[]){
System.out.println("Are you a boy or a girl?");
Scanner lalala = new Scanner(System.in);
String text = lalala.nextLine();
if (text == "boy"){
System.out.println("You are a boy.");
}
else{
System.out.println("You are a girl.");
}
}
}
Thanks a lot~
Use
text.equals("boy")
instead of
if (text == "boy")
The reason is, In Java, == always just compares two references (for non-primitives, that is) - i.e. it tests whether the two operands refer to the same object.
However, the equals method can be overridden - so two distinct objects can still be equal.
But the better option is to use equalsIgnoreCase, the reason is that user may enter boy or BOY or Boy or any other combination. equalsIgnoreCase method just ignores the case and compares two strings.
text.equalsIgnoreCase("boy")
Side note. Input other than Boy will get you to print Girl. Put one more extra if condition. Also before going for comparision, being on safer side trim the string for blank spaces.
Use
text.equals("boy")
or
text.equalIgnoreCase("boy")
if its not case sensitive.
instead of
if (text == "boy")
Your coade will be something like this
import java.util.Scanner;
class apples{
public static void main(String args[]){
System.out.println("Are you a boy or a girl?");
Scanner lalala = new Scanner(System.in);
String text = lalala.nextLine();
text = text.trim();
if (text.equalIgnoreCase("Boy")){
System.out.println("You are a boy.");
}
else if(text.equalIgnoreCase("Girl")){
System.out.println("You are a girl.");
} else {
System.out.println("Invalid Gender");
}
}
}
use text.equalsIgnoreCase("boy")
if you think about case also it will check content of string is same.
== is used to check for object equality.
text.equals("boy") is ok if you not consider about case.
Please use equals method for string like : if (text.equals("boy")){
== is not for string contents equality check in java
== tests if object identity is the same. When you have two string objects containing the same value this object identity won't be equal. Use the equals function to test for logical equivalence.
Your if else is wrong. Use only if here. Otherwise,
If I entered "some text", the output will be "You are a girl"
Like this
if (text.equalsIgnoreCase("boy")) {
System.out.println("You are a boy.");
}
else if (text.equalsIgnoreCase("girl")) {
System.out.println("You are a girl.");
} else {
System.out.println("You are neither a boy nor a girl");
}
It would be best to use the .equals method instead of == boy
For your other question regarding assert:
Java has an assert function. This is the sample code.
assert(x <= 50): "Not Accepted";
This code checks if integer x is greater than 50. You will see the output at the console.
This question already has answers here:
How do I compare strings in Java?
(23 answers)
Closed 4 years ago.
I am writing a program that should close the console if the user input the String "end'.
The program always performs the else loop even if the user inputs "end". I'm wondering why the program is not getting into the if part of the loop and shutting down.
Scanner scan = new Scanner(System.in);
while(true)
{
String num = scan.nextLine();
if(num == "end")
{
System.exit(0);
}
else
{
System.out.println("hi");
}
}
You're using == instead of "end".equals(num)
Don't use == for equality of string as it compares the objects not the string itself.
Use num.equals("end") or num.equalsIgnoreCase("end") if you want to be able to type end or END
I would not use "end".equals(num), although considered better from a performance perspective in most cases, it does not clearly state the business requirement and it is more important to make it more readable.
But be aware of num being null, if that is possible, num.quals("end") could throw an exception and you should write if (num!=null && num.equals("end")) { ... }
Note that "end".equals(num) does not need the null check, but I still believe this is not very readable, so I would go with if (num!=null && num.equals("end")) { ... }
For testing equality between strings, you should use equals() instead.
if(a.equals(b)) and so on.
This should help you out: http://leepoint.net/notes-java/data/expressions/22compareobjects.html
In Java, you test equality of strings with:
string1.equals(string2);
So in this case, it would be:
num.equals("end");
Or to avoid an exception of type NullPointerException:
"end".equals(num);
num refers to the object, so num == "end" should never be. You want num.equals("end")
Please do not use comparison (==) operator when comparing objects in Java. Use equals(Object) instead.