Develop Reference

same string but it return false while using equals method with two StringBuilder objects [duplicate]

class strb
{
static public void main(String...string)
{
StringBuilder s1 = new StringBuilder("Test");
StringBuilder s2 = new StringBuilder("Test");
System.out.println(s1); // output: Test
System.out.println(s2); // Test
System.out.println(s1==s2); // false
System.out.println(s1.equals(s2)); //Line 1 output: false
System.out.println(s1.toString()==s2.toString()); //Line 2 output: false
}
}
Just have a quick question on .equals.
Regardless of the object content, does .equals return true only if both the object references point to the same object ?
EDIT : Now I understand the part about the .equals but why does Line 2 not return true ?
EDIT : I believe == looks at the reference variable's address and so s1 and s2 cannot be equal.correct me if my assumption is not right

Yes, StringBuilder does not override Object's .equals() function, which means the two object references are not the same and the result is false.
For StringBuilder, you could use s1.toString().equals(s2.toString())
For your edit, you're calling the == operator on two different String objects. The == operator will return false because the objects are different. To compare Strings, you need to use String.equals() or String.equalsIgnoreCase()
It's the same problem you were having earlier

The StringBuilder class does not provide an overriden equals() method. As such, when that method is called on an instance of StringBuilder, the Object class implementation of the method is executed, since StringBuilder extends Object.
The source code for that is
public boolean equals(Object obj) {
return (this == obj);
}
Which simply compares reference equality.

The default implementation of .equals for the Object class is as you mentioned.
Other classes can override this behavior. StringBuilder is not one of them.
String is one of them, which overrides it to ensure that the String representations of both objects result in the same sequence of characters. String API
Refer to the documentation for the specific object in question.

Check the contract of equals method:
it must be consistent (if the objects are not modified, then it must keep returning the same value).
That's why StringBuilder does not override it regardless of its content.
Let's take example above.
StringBuilder s1 = new StringBuilder("Test");
StringBuilder s2 = new StringBuilder("Test");
Maybe, to you it is expected that s1.equals(s2) returns true due to current run time values.
But what about if you change add line:
s1.append("abc");
Then s1 and s2 will have different string contents and s1.equals(s2) is expected to be false. But it is in contradiction with consistency.

for your first answer check #abmitchell 's Answer
And for your Edit:
In Java, String is an object and we can't compare objects for value equality by using ==
== is used for comparing primitives values or object references.
To compare Object values we use equals() in Java

StringBuilder class doesn't have the implementation of equals() method like one in the String class.
So it executes default Object class functionality, which again checks only for address equivalency, which is not same in this case. so it prints false.
Note 1: You can use == operator on objects also, but it simply checks if the address of both the objects are same or not.
Note 2: == operator plays good role in comparing two String objects created in string constant pool, to find out if it is really creating a new object in string constant pool or not.

StringBuilder and StringBuffer not override the equals function of Object class.but string override the equals method.
the function of Object is this
public boolean equals(Object obj) {
return (this == obj);
}
you could write your code like this.
System.out.println(s1.toString() == s2.toString());
System.out.println(s1.toString().equals(s2.toString()));

Agreed with both of the above responses, but I'm worth noting to actually compare contents you can do:
System.out.println(s1.toString().equals(s2.toString())); //Line 1

Since StringBuilder does not have a equals method, it is best to first change the StringBuilder to String and then check equality.
Eg - sb1.toString().equals(sb2.toString()).
Note: The == operator or sb1 == sb2 will never work here because the two StringBuilder(s) are completely different objects.

As others have said, StringBuilder does not override equals. However, it does implement Comparable (since Java 11). Therefore, you can check s1.compareTo(s2) == 0 to find out if the string representations are equal.

functioning of (== ) in terms of hashCode

String s1="abc";
String s2=new String("abc");
when we compare the both
s1==s2; it return false
and when we compare it
with s1.hashCode()==s2.hashCode it return true
i know (==) checks reference id's .Does it returning true in above comparison because above hashCode are saved to same bucket??Please give me explanation

Don't forget that your hash codes are primitive integers, and comparing primitives using == will compare their values, not their references (since primitives don't have references)
Hence two strings having the same contents will yield the same hash code, and a comparison via == is perfectly valid.
The concept of a bucket is only valid when you put an object into a hashed collection (e.g. a HashSet). The value of the hash code dictates which bucket the object goes into. The hash codes themselves aren't stored.

First comparison fails because they are two different objects. Second comparison works because they are comparing the output of the hashCode() function, which produces the same value for both.

When comparing objects using the equality operator (==) you are testing the equality of the Object Reference. Object References are only equal if they are the same object.
What is usually meant is the equality of the concept/information the object encapsulates. And this is best determined by the equal(Object obj) method.
For String this would be:
s1.equals(s2);

When you compare the two Strings using s1==s2, you are comparing references. Since just before that line you created a new String object by copying the old one, both references refer to different objects, even though they contain the same value.
On the other hand, what the hashCode method does, depends on how it is implemented. The specification says that it has to comply with the equals function (whose behavior also depends on how it is implemented in the concrete type): If two objects are equal according to the equals method, they must return the same hash code. Since equals for String compares the values (not the references), both hashCode calls have to return the same value. And since hashCode returns an int, which is a primitive data type, your == comparison actually compares values and therefore works as expected.
Note: If you had done the same experiment for a class different from String that doesn't override the equals and hashCode methods (which is perfectly valid; it's just another way of defining equality on your type of object), it would have returned false for the hashCode comparison, too:
public class MyObject {
private int value;
public MyObject(MyObject toCopy) {
this.value = toCopy.value;
}
public MyObject(int value) {
this.value = value;
}
public static void main(String[] args) {
MyObject s1 = new MyObject(0);
MyObject s2 = new MyObject(s1);
System.out.println(s1 == s2); // false
System.out.println(s1.hashCode() == s2.hashCode()); // false
}
}
It all depends on how the programmer of the class defines which objects to consider equal.

Why == operator is not working here, but equals work in java

I have written the code to check whether the given string is palindrome or not. But here I didn't create any String object explicitly. When we don't create explicitly, "==" also should work to compare the strings. But here I am not getting correct output if I use ==. For the clarity in my question, I have given another code also below
Code 1:Here. "==" is not working.
class Palindrome
{
public static void main(String[] args)
{
StringBuffer sb1=new StringBuffer();
sb1.append("anna");
String s1=sb1.toString();
StringBuffer sb2=new StringBuffer();
sb2=sb1.reverse();
String s2=sb2.toString();
if(s1.equals(s2))
{
System.out.println("The given String is a Palindrome");
}
else
System.out.println("Not a Palindrome");
}
}
Code 2: Here == works
class Stringdemo
{
public static void main(String[] args)
{
String str1="hello";
String str2="hello";
if(str1==str2)
{
System.out.println("both strings are same");
}
else
{
System.out.println("both strings are not Same");
}
}
}

Using StringBuilder/Buffer.toString() will create a new instance hence why equals() works, but == doesn't. In the case of using string literals, any string that is a string literal will be added to the constants pool of the class, and hence string1 and string2 will merely point to that same reference within the constant pool for "hello", ergo == says true because they are the same reference.

"==" compares object reference values where "equals()" compares object contents. "==" is only really useful for comparing primitives.
When you call "sb2.toString()" it creates a new object which has a new reference value.

sb2.toString(); creates a new String. == won't work here
it will however work if you use sb2.toString().intern() instead

== checks for object identity equality (i.e., same memory address).
.equals() checks for for equality of the value.
When you create a new string literal:
String str1="hello";
String str2="hello";
Java calls the String.intern() method to intern the new String, but will first check to see if that same String value already exists, and if it does it will return that (i.e., will not create a new instance). See the Javadoc comments on String.intern() for more details:
When the intern method is invoked, if the pool already contains a string equal to this String object as determined by the equals(Object) method, then the string from the pool is returned. Otherwise, this String object is added to the pool and a reference to this String object is returned.
This is why the second example works, but the StringBuilder example does not since calling sb2.toString creates a new instance, bypassing intern().
To be safe, you are better off using .equals() for any non-primative objects.

== operator just checks the object's reference, while equals() compares the strings for its characters seperately.
Since String are immutable objects therefore it shares the reference of objects if it is created already in memory.
Therefore,
String str1 = "hello"; //and
String str2 = "hello"; //are equal
Because both str1 and str2 shares the same reference of "hello".

String compare strange bug

I have strange bug in my code. I have variable type and when I load that from class Settings ( implements persistable ) it has value "CPU 21" (type="CPU 21"), but when I try
if(type=="CPU 21") the condition is false. How is that possible ?

It's not a bug at all. It's the way that == works.
For reference types (such as string), "==" will always compare the references directly. If you have two references which refer to separate objects with equal content, then "==" will evaluate to false. Use equals for value equality.
Note that it's easy to fool yourself by using String literals, which are interned:
String x = "CPU 21";
String y = "CPU 21";
boolean b = (x == y);
Now b is true because the values of x and y - the references are actually the same. They're referreing to the same String object. Compare that with this:
String x = new String("CPU 21");
String y = new String("CPU 21");
boolean b1 = (x == y);
boolean b2 = x.equals(y);
Now b1 is false because x and y refer to distinct String objects, but b2 is true because String.equals(String) compares the contents of the strings in question (i.e. the character sequences represented by the strings).

You have to use .equals() to compare string content:
if (type.equals("CPU 21")) ...
This is because the == operator only compares two references to see whether they are pointing to the same object or not. Since your string type is not the same object as the literal "CPU 21", the comparison is false. The .equals() method actually checks the strings themselves.
You may also find that people often write this the other way around:
if ("CPU 21".equals(type)) ...
This means the same thing, but will not throw an exception if type happens to be null.

What is difference between equals() and == ?
Explanation : 1
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.
Explanation : 2
By definintion, the objects are all created on the heap. When you create an object, say,
Object ob1 = new SomeObject();
Object ob2 = new SomeObject();
We have 2 objects with exactly the same contents, lets assume. We also have 2 references, lets say ob1 is at address 0x1234 and ob2 is at address 0x2345. Though the contents of the objects are the same, the references differ.
Using == compares the references. Though the objects, ob1 and ob2 are same internally, they differ on using this operation as we comare references. ob1 at address 0x1234 is compared with ob2 at address 0x2345. Hence, this comparison would fail.
object.equals() on the other hand compares the values. Hence, the comparison between ob1 and ob2 would pass. Note that the equals method should be explicitly overridden for this comparison to succeed.

you need to use equals(); to compare String values
replace
if(type=="CPU 21")
with
if("CPU 21".equals(type))
== will compare two reference variable. while equals will check the object's equality
Also See
java-string-equals-versus ==

String.equals versus == [duplicate]

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.....