This question already has answers here:
How do I compare strings in Java?
(23 answers)
Closed 9 years ago.
This code separates a string into tokens and stores them in an array of strings, and then compares a variable with the first home ... why isn't it working?
public static void main(String...aArguments) throws IOException {
String usuario = "Jorman";
String password = "14988611";
String strDatos = "Jorman 14988611";
StringTokenizer tokens = new StringTokenizer(strDatos, " ");
int nDatos = tokens.countTokens();
String[] datos = new String[nDatos];
int i = 0;
while (tokens.hasMoreTokens()) {
String str = tokens.nextToken();
datos[i] = str;
i++;
}
//System.out.println (usuario);
if ((datos[0] == usuario)) {
System.out.println("WORKING");
}
}
Use the string.equals(Object other) function to compare strings, not the == operator.
The function checks the actual contents of the string, the == operator checks whether the references to the objects are equal. Note that string constants are usually "interned" such that two constants with the same value can actually be compared with ==, but it's better not to rely on that.
if (usuario.equals(datos[0])) {
...
}
NB: the compare is done on 'usuario' because that's guaranteed non-null in your code, although you should still check that you've actually got some tokens in the datos array otherwise you'll get an array-out-of-bounds exception.
Meet Jorman
Jorman is a successful businessman and has 2 houses.
But others don't know that.
Is it the same Jorman?
When you ask neighbours from either Madison or Burke streets, this is the only thing they can say:
Using the residence alone, it's tough to confirm that it's the same Jorman. Since they're 2 different addresses, it's just natural to assume that those are 2 different persons.
That's how the operator == behaves. So it will say that datos[0]==usuario is false, because it only compares the addresses.
An Investigator to the Rescue
What if we sent an investigator? We know that it's the same Jorman, but we need to prove it. Our detective will look closely at all physical aspects. With thorough inquiry, the agent will be able to conclude whether it's the same person or not. Let's see it happen in Java terms.
Here's the source code of String's equals() method:
It compares the Strings character by character, in order to come to a conclusion that they are indeed equal.
That's how the String equals method behaves. So datos[0].equals(usuario) will return true, because it performs a logical comparison.
It's good to notice that in some cases use of "==" operator can lead to the expected result, because the way how java handles strings - string literals are interned (see String.intern()) during compilation - so when you write for example "hello world" in two classes and compare those strings with "==" you could get result: true, which is expected according to specification; when you compare same strings (if they have same value) when the first one is string literal (ie. defined through "i am string literal") and second is constructed during runtime ie. with "new" keyword like new String("i am string literal"), the == (equality) operator returns false, because both of them are different instances of the String class.
Only right way is using .equals() -> datos[0].equals(usuario). == says only if two objects are the same instance of object (ie. have same memory address)
Update: 01.04.2013 I updated this post due comments below which are somehow right. Originally I declared that interning (String.intern) is side effect of JVM optimization. Although it certainly save memory resources (which was what i meant by "optimization") it is mainly feature of language
The == operator checks if the two references point to the same object or not.
.equals() checks for the actual string content (value).
Note that the .equals() method belongs to class Object (super class of all classes). You need to override it as per you class requirement, but for String it is already implemented and it checks whether two strings have the same value or not.
Case1)
String s1 = "Stack Overflow";
String s2 = "Stack Overflow";
s1 == s1; // true
s1.equals(s2); // true
Reason: String literals created without null are stored in the string pool in the permgen area of the heap. So both s1 and s2 point to the same object in the pool.
Case2)
String s1 = new String("Stack Overflow");
String s2 = new String("Stack Overflow");
s1 == s2; // false
s1.equals(s2); // true
Reason: If you create a String object using the `new` keyword a separate space is allocated to it on the heap.
equals() function is a method of Object class which should be overridden by programmer. String class overrides it to check if two strings are equal i.e. in content and not reference.
== operator checks if the references of both the objects are the same.
Consider the programs
String abc = "Awesome" ;
String xyz = abc;
if(abc == xyz)
System.out.println("Refers to same string");
Here the abc and xyz, both refer to same String "Awesome". Hence the expression (abc == xyz) is true.
String abc = "Hello World";
String xyz = "Hello World";
if(abc == xyz)
System.out.println("Refers to same string");
else
System.out.println("Refers to different strings");
if(abc.equals(xyz))
System.out.prinln("Contents of both strings are same");
else
System.out.prinln("Contents of strings are different");
Here abc and xyz are two different strings with the same content "Hello World". Hence here the expression (abc == xyz) is false where as (abc.equals(xyz)) is true.
Hope you understood the difference between == and <Object>.equals()
Thanks.
== tests for reference equality.
.equals() tests for value equality.
Consequently, if you actually want to test whether two strings have the same value you should use .equals() (except in a few situations where you can guarantee that two strings with the same value will be represented by the same object eg: String interning).
== is for testing whether two strings are the same Object.
// These two have the same value
new String("test").equals("test") ==> true
// ... but they are not the same object
new String("test") == "test" ==> false
// ... neither are these
new String("test") == new String("test") ==> false
// ... but these are because literals are interned by
// the compiler and thus refer to the same object
"test" == "test" ==> true
// concatenation of string literals happens at compile time resulting in same objects
"test" == "te" + "st" ==> true
// but .substring() is invoked at runtime, generating distinct objects
"test" == "!test".substring(1) ==> false
It is important to note that == is much cheaper than equals() (a single pointer comparision instead of a loop), thus, in situations where it is applicable (i.e. you can guarantee that you are only dealing with interned strings) it can present an important performance improvement. However, these situations are rare.
Instead of
datos[0] == usuario
use
datos[0].equals(usuario)
== compares the reference of the variable where .equals() compares the values which is what you want.
Let's analyze the following Java, to understand the identity and equality of Strings:
public static void testEquality(){
String str1 = "Hello world.";
String str2 = "Hello world.";
if (str1 == str2)
System.out.print("str1 == str2\n");
else
System.out.print("str1 != str2\n");
if(str1.equals(str2))
System.out.print("str1 equals to str2\n");
else
System.out.print("str1 doesn't equal to str2\n");
String str3 = new String("Hello world.");
String str4 = new String("Hello world.");
if (str3 == str4)
System.out.print("str3 == str4\n");
else
System.out.print("str3 != str4\n");
if(str3.equals(str4))
System.out.print("str3 equals to str4\n");
else
System.out.print("str3 doesn't equal to str4\n");
}
When the first line of code String str1 = "Hello world." executes, a string \Hello world."
is created, and the variable str1 refers to it. Another string "Hello world." will not be created again when the next line of code executes because of optimization. The variable str2 also refers to the existing ""Hello world.".
The operator == checks identity of two objects (whether two variables refer to same object). Since str1 and str2 refer to same string in memory, they are identical to each other. The method equals checks equality of two objects (whether two objects have same content). Of course, the content of str1 and str2 are same.
When code String str3 = new String("Hello world.") executes, a new instance of string with content "Hello world." is created, and it is referred to by the variable str3. And then another instance of string with content "Hello world." is created again, and referred to by
str4. Since str3 and str4 refer to two different instances, they are not identical, but their
content are same.
Therefore, the output contains four lines:
Str1 == str2
Str1 equals str2
Str3! = str4
Str3 equals str4
You should use string equals to compare two strings for equality, not operator == which just compares the references.
It will also work if you call intern() on the string before inserting it into the array.
Interned strings are reference-equal (==) if and only if they are value-equal (equals().)
public static void main (String... aArguments) throws IOException {
String usuario = "Jorman";
String password = "14988611";
String strDatos="Jorman 14988611";
StringTokenizer tokens=new StringTokenizer(strDatos, " ");
int nDatos=tokens.countTokens();
String[] datos=new String[nDatos];
int i=0;
while(tokens.hasMoreTokens()) {
String str=tokens.nextToken();
datos[i]= str.intern();
i++;
}
//System.out.println (usuario);
if(datos[0]==usuario) {
System.out.println ("WORKING");
}
Generally .equals is used for Object comparison, where you want to verify if two Objects have an identical value.
== for reference comparison (are the two Objects the same Object on the heap) & to check if the Object is null. It is also used to compare the values of primitive types.
== operator compares the reference of an object in Java. You can use string's equals method .
String s = "Test";
if(s.equals("Test"))
{
System.out.println("Equal");
}
If you are going to compare any assigned value of the string i.e. primitive string, both "==" and .equals will work, but for the new string object you should use only .equals, and here "==" will not work.
Example:
String a = "name";
String b = "name";
if(a == b) and (a.equals(b)) will return true.
But
String a = new String("a");
In this case if(a == b) will return false
So it's better to use the .equals operator...
The == operator is a simple comparison of values.
For object references the (values) are the (references). So x == y returns true if x and y reference the same object.
I know this is an old question but here's how I look at it (I find very useful):
Technical explanations
In Java, all variables are either primitive types or references.
(If you need to know what a reference is: "Object variables" are just pointers to objects. So with Object something = ..., something is really an address in memory (a number).)
== compares the exact values. So it compares if the primitive values are the same, or if the references (addresses) are the same. That's why == often doesn't work on Strings; Strings are objects, and doing == on two string variables just compares if the address is same in memory, as others have pointed out. .equals() calls the comparison method of objects, which will compare the actual objects pointed by the references. In the case of Strings, it compares each character to see if they're equal.
The interesting part:
So why does == sometimes return true for Strings? Note that Strings are immutable. In your code, if you do
String foo = "hi";
String bar = "hi";
Since strings are immutable (when you call .trim() or something, it produces a new string, not modifying the original object pointed to in memory), you don't really need two different String("hi") objects. If the compiler is smart, the bytecode will read to only generate one String("hi") object. So if you do
if (foo == bar) ...
right after, they're pointing to the same object, and will return true. But you rarely intend this. Instead, you're asking for user input, which is creating new strings at different parts of memory, etc. etc.
Note: If you do something like baz = new String(bar) the compiler may still figure out they're the same thing. But the main point is when the compiler sees literal strings, it can easily optimize same strings.
I don't know how it works in runtime, but I assume the JVM doesn't keep a list of "live strings" and check if a same string exists. (eg if you read a line of input twice, and the user enters the same input twice, it won't check if the second input string is the same as the first, and point them to the same memory). It'd save a bit of heap memory, but it's so negligible the overhead isn't worth it. Again, the point is it's easy for the compiler to optimize literal strings.
There you have it... a gritty explanation for == vs. .equals() and why it seems random.
#Melkhiah66 You can use equals method instead of '==' method to check the equality.
If you use intern() then it checks whether the object is in pool if present then returns
equal else unequal. equals method internally uses hashcode and gets you the required result.
public class Demo
{
public static void main(String[] args)
{
String str1 = "Jorman 14988611";
String str2 = new StringBuffer("Jorman").append(" 14988611").toString();
String str3 = str2.intern();
System.out.println("str1 == str2 " + (str1 == str2)); //gives false
System.out.println("str1 == str3 " + (str1 == str3)); //gives true
System.out.println("str1 equals str2 " + (str1.equals(str2))); //gives true
System.out.println("str1 equals str3 " + (str1.equals(str3))); //gives true
}
}
The .equals() will check if the two strings have the same value and return the boolean value where as the == operator checks to see if the two strings are the same object.
Someone said on a post higher up that == is used for int and for checking nulls.
It may also be used to check for Boolean operations and char types.
Be very careful though and double check that you are using a char and not a String.
for example
String strType = "a";
char charType = 'a';
for strings you would then check
This would be correct
if(strType.equals("a")
do something
but
if(charType.equals('a')
do something else
would be incorrect, you would need to do the following
if(charType == 'a')
do something else
a==b
Compares references, not values. The use of == with object references is generally limited to the following:
Comparing to see if a reference is null.
Comparing two enum values. This works because there is only one object for each enum constant.
You want to know if two references are to the same object
"a".equals("b")
Compares values for equality. Because this method is defined in the Object class, from which all other classes are derived, it's automatically defined for every class. However, it doesn't perform an intelligent comparison for most classes unless the class overrides it. It has been defined in a meaningful way for most Java core classes. If it's not defined for a (user) class, it behaves the same as ==.
Use Split rather than tokenizer,it will surely provide u exact output
for E.g:
string name="Harry";
string salary="25000";
string namsal="Harry 25000";
string[] s=namsal.split(" ");
for(int i=0;i<s.length;i++)
{
System.out.println(s[i]);
}
if(s[0].equals("Harry"))
{
System.out.println("Task Complete");
}
After this I am sure you will get better results.....
This question already has answers here:
How do I compare strings in Java?
(23 answers)
Closed 8 years ago.
I'm in little trouble. The problem is when I'm trying to compare 2 strings(type String) operator '==' returns FALSE, but actually strings are equal.
Here's the code with its problem:
//before the following code I filled the "LinkedList <String> command" and there is
//a node with value of args[0]
String deal="";
Iterator it = commands.listIterator();
if(it.hasNext() == true)
{
if(it.next() == args[0])
{
deal += it.next();
it.hasNext();
break;
}
}
Thank You!!!
To compare two strings u should use the method equals() or equalsIgnoreCase().
in your case:
if(it.next().equals(args[0]))
the operator == returns true if the two object are the same object, same address in memory.
You use .equals when comparing two strings. So use
(it.next()).equals(args[0])
You have to use .equals method:
String deal="";
Iterator it = commands.listIterator();
if(it.hasNext() == true)
{
String next = it.next();
if(next.equals(args[0]))
{
deal += next;
break;
}
}
Be careful, .next() returns the value once and move its internal cursor to the next value.
The == cannot be used for String because the == is true if the same object instance is on both sides. The same string content can be in many String instances.
There are two ways of comparing strings.
Comparing the value of the strings (achieved using .equals ).
Comparing the actual object (achieved using == operator).
In your code you are comparing the references referred by it.next() & args[0]whereas you should compare the value of the two using it.next().equals(args[0]).
if you use == to compare two int values, then it is compare the two values, because int is primitive data type. If you use "==" to compare String object, it is check whether both String reference are referring the same String object or not. It do not consider values of the String objects.
If you want to compare values of String objects you have to use equals() of the String class. This method is comparing content of both String objects.
This question already has answers here:
How do I compare strings in Java?
(23 answers)
What's the difference between ".equals" and "=="? [duplicate]
(11 answers)
Closed 9 years ago.
This is the piece of code I used from the SCJP book.
I understand that == compares the memory location of the objects to find the equality and .equals compares the hashcode to determine the equality.
My question is in the below snippet, in the overrided equals method we compare :
(((Moof)o).getMoofValue()) == this.getMoofValue()
in the above code, does the == compare the memory location of the string value? If it does then it should be false. But it returns true. How does the == work here?
public class EqualsTest {
public static void main(String args[]){
Moof one = new Moof("a");
Moof two = new Moof("a");
if(one.equals(two)){
System.out.println("Equal");
}
else{
System.out.println("Not Equal");
}
}
}
public class Moof {
private String moofValue;
public Moof(String i){
this.moofValue = i;
}
public String getMoofValue(){
return this.moofValue;
}
public boolean equals(Object o){
if((o instanceof Moof) && (((Moof)o).getMoofValue()) == this.getMoofValue()){
return true;
}
return false;
}
}
(Moof)o).getMoofValue()) == this.getMoofValue())
here..You are still comparing references to the same String object in the String pool. Because String literals are interned automatically.
Edit :
public class TestClass {
String s;
TestClass(String s) {
this.s = s;
}
public static void main(String[] args) {
String s1 = new String("a");
String s2 = new String("a");
System.out.println(s1 == s2);
TestClass t1 = new TestClass("a");
TestClass t2 = new TestClass("a");
System.out.println(t1 == t2); // here you are comparing t1 with t2, not t1.s with t2.s so you get false...
System.out.println(t1.s == t2.s); // t1.s and t2.s refer to the same "a", so you get true.
TestClass t3 = new TestClass(s1);
TestClass t4 = new TestClass(s2);
System.out.println(t3.s == t4.s); // false because s1 and s2 are 2 different references.
}
}
O/P :
false
false
true
false
The "a" string literals are interned. Each "a" is the same string in memory, so they compare equal with ==. Note that you can't rely on this for strings in general. If you did the following:
Moof two = new Moof("aa".substring(1));
the strings would be considered != to each other.
You are right that == compares if the references point to the same object. So, two different objects would evaluate to false with == even if equals() would evaluate to true.
S the following code
(((Moof)o).getMoofValue()) == this.getMoofValue()
checks if the objects are the same. However, in Java two equal string constants that are known at compile time must reference the same string object. Looking at
Moof one = new Moof("a");
Moof two = new Moof("a");
We could view this as
String theValue = "a";
Moof one = new Moof(theValue);
Moof two = new Moof(theValue);
However, making a new string must return a new object so you can try
Moof one = new Moof(new String("a"));
Moof two = new Moof(new String)"a");
which would cause the equality to evaluate to false. More specific example below:
String a1 = "a";
String a2 = "a";
String a3 = new String("a");
System.out.println(a1 == a2); // true
System.out.println(a1 == a3); // false
Equals does normally not use hashCode to check for equality but it could be used to make a quick check if more comparisons are needed. The hashCode is a numerical indicator of the object, which if you implemented a PersonId class could be a numerical value of the country phone prefix and age of the person. So a 27 year old person from Sweden would get hash code 4627. Now, when checking if two PersonId refers to the same person you might have to do a lot of comparisons, but if the hashCode is not the same then no more comparisons are needed (since either the country or age is different and the PersonIds must refer to different persons).
The hashCode is used in structures such as HashMap as a value for knowing which "bucket" to store the object in.
I understand that == compares the memory location of the objects to find the equality and .equals compares the hashcode to determine the equality.
No it doesn't. .equals() does whatever the guy who wrote it wrote. In the case of Object.equals(), it returns the result of ==. hashCode() has nothing to do with it.
in the above code, does the == compare the memory location of the string value?
Yes, you've already said that.
If it does then it should be false. But it returns true. How does the == work here?
Because both occurrences of "a" are pooled to the same value, with the same address.
This question already has answers here:
How do I compare strings in Java?
(23 answers)
Closed 7 years ago.
This seems to be pretty simple, but I have been stucked here for a couple of hours.
I have a doubt when you have to compare two Strings in Java.
if I just do something like this:
String var1 = "hello";
String var2 = "hello";
and then compare these two words in another function, the result will clearly be true.
But the problem is when I have to compare two words that come from an input. Here is my code:
import java.util.Scanner;
public class Compare{
public static void main(String[] args){
Scanner Scanner = new Scanner (System.in);
System.out.println("Enter first word: ");
String var1 = Scanner.nextLine();
System.out.println("Enter second word: ");
String var2 = Scanner.nextLine();
if (same (var1, var2))
System.out.println("Yes");
else
System.out.println("No");
}
public static boolean same (String var1, String var2){
if (var1 == var2)
return true;
else
return false;
}
}
I have tried several times (clearly entering the same word) and the result is always False.
I don't know why this happens. What am I missing?
This is my first time in Java. I will appreciate any kind of help. Thanks
You should change
if (var1 == var2)
{
return true;
}
else
{
return false;
}
to
if (var1.equals(var2))
{
return true;
}
else
{
return false;
}
See this answer for the difference between the two
To be more accurate, with Strings in Java sometimes you can use == instead of .equals, if your string has been interned. Remember that == always compares the object references, not the contents of the object. Interning a String means that you will get the same object reference back and this is why == works with interned Strings.
Please read the Javadoc here to understand this more clearly:
String.intern()
In Java the == is a reference equality operator.
It works with the following.
String var1 = "hello";
String var2 = "hello";
boolean cmp = var1 == var2;
just because they are string literals and they are allocated in the same place inside the string table, so both variables point to the same string.
If you are fetching data from another source the strings are dynamically allocated, hence you should use the var1.equals(var2) (and you should ALWAYS use that one when comparing two objects).
Instead of if (same (var1, var2)) use if (v1.equals(v2)). No need to create a new method to compare two Strings. That's what equals() does.
== is used to compares references, not the contents of each String object.
The equality operator(==) checks the refernce of string first then checks value of string.
While equals method checks the value first.
So,in this case equals method should be used instead of equality operator.
String s="hello";
String s1="hello";
String s3=new String("hello")
In the above code snippet if you use If(s==s1){System.out.print("Equal");}it would print equal.But if you check If(s==s3){System.out.print("unqual");}it wouldn't print unequal.
so,you can see that even strings s and s3 are equal,output is wrong.Therefore,in this scenario like program in question
Equals method must be used.
var1 == var2
sometimes works because VM allocates the same memory both the variables for memory optimization and thus having same reference. That cannot be always the case so it's better to use
var1.equals(var2)
If you want to compare their values and doesnt care about reference.
This question already has answers here:
Closed 11 years ago.
Possible Duplicate:
Java string comparison?
I was trying to do this:
boolean exit = false;
while(exit==false && convStoreIndex<convStoreLength) {
if(conversionStore[convStoreIndex].getInUnit()==inUnit) {
indexCount++;
exit=true;
}
convStoreIndex++;
}
but the if-condition never went true, even if the two Strings were the same(checked this in the debugger).
so I added some lines:
boolean exit = false;
while(exit==false && convStoreIndex<convStoreLength) {
Log.v("conversionStore["+String.valueOf(convStoreIndex)+"]", conversionStore[convStoreIndex].getInUnit()+"|"+inUnit);
String cs = conversionStore[convStoreIndex].getInUnit();
String iu = inUnit;
Log.v("cs", cs);
Log.v("iu", iu);
Log.v("Ergebnis(cs==iu)", String.valueOf(cs==iu));
if(conversionStore[convStoreIndex].getInUnit()==inUnit) {
indexCount++;
exit=true;
}
convStoreIndex++;
}
and here is the extract from LogCat:
09-15 11:07:14.525: VERBOSE/cs(585): kg
09-15 11:07:16.148: VERBOSE/iu(585): kg
09-15 11:07:17.687: VERBOSE/Ergebnis(cs==iu)(585): false
the class of conversionStore:
class ConversionStore {
private String inUnit;
[...]
public String getInUnit() {
return inUnit;
}
}
Who is going crazy, java or me?
Don't use == to compare String objects, use .equals():
if(conversionStore[convStoreIndex].getInUnit().equals(inUnit)) {
To compare Strings for equality, don't use ==. The == operator checks
to see if two objects are exactly the same object. Two strings may be
different objects, but have the same value (have exactly the same
characters in them). Use the .equals() method to compare strings for
equality.
Straight from the first link Google provided when searching "Java string comparison"...
Please use String.equals() to compare by string content instead of reference identity.
You are comparing your Strings using == : cs==iu.
But this will return true only if both Strings are actually the same object. This is not the case here: you have two distinct instances of String that contain the same value.
You should use String.compareTo(String).
use .equals(); method instread of ==
Becase ::
They both differ very much in their significance. equals() method is present in the java.lang.Object class and it is expected to check for the equivalence of the state of objects! That means, the contents of the objects. Whereas the '==' operator is expected to check the actual object instances are same or not.
For example, lets say, you have two String objects and they are being pointed by two different reference variables s1 and s2.
s1 = new String("abc");
s2 = new String("abc");
Now, if you use the "equals()" method to check for their equivalence as
if(s1.equals(s2))
System.out.println("s1.equals(s2) is TRUE");
else
System.out.println("s1.equals(s2) is FALSE");
You will get the output as TRUE as the 'equals()' method check for the content equivality.
Lets check the '==' operator..
if(s1==s2)
System.out.printlln("s1==s2 is TRUE");
else
System.out.println("s1==s2 is FALSE");
Now you will get the FALSE as output because both s1 and s2 are pointing to two different objects even though both of them share the same string content. It is because of 'new String()' everytime a new object is created.
Try running the program without 'new String' and just with
String s1 = "abc";
String s2 = "abc";
You will get TRUE for both the tests.
Reson::
After the execution of String str1 = “Hello World!”; the JVM adds the string “Hello World!” to the string pool and on next line of the code, it encounters String str2 =”Hello World!”; in this case the JVM already knows that this string is already there in pool, so it does not create a new string. So both str1 and str2 points to the same string means they have same references. However, str3 is created at runtime so it refers to different string.
Two things:
You don't need to write "boolean==false" you can just write "!boolean" so in your example that would be:
while(!exit && convStoreIndex<convStoreLength) {
Second thing:
to compare two Strings use the String.equals(String x) method, so that would be:
if(conversionStore[convStoreIndex].getInUnit().equals(inUnit)) {