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What use does the == operator have for String?

In Java, if one is to check if two Strings are equal, in the sense that their values are the same, he/she needs to use the equals method. E.g. :
String foo = "foo";
String bar = "bar";
if(foo.equals(bar)) { /* do stuff */ }
And if one wants to check for reference equality he needs to use the == operator on the two strings.
if( foo == bar ) { /* do stuff */ }
So my question is does the == operator have it's use for the String class ? Why would one want to compare String references ?
Edit:
What I am not asking : How to compare strings ? How does the == work ? How does the equals method work?
What I am asking is what uses does the == operator have for String class in Java ? What is the justification of not overloading it, so that it does a deep comparison ?

Imagine a thread-safe Queue<String> acting as a communication channel between a producer thread and a consumer thread. It seems perfectly reasonable to use a special String to indicate termination.
// Deliberate use of `new` to make sure JVM does not re-use a cached "EOT".
private static final String EOT = new String("EOT");
...
// Signal we're done.
queue.put(EOT);
// Meanwhile at the consumer end of the queue.
String got = queue.get();
if ( got == EOT ) {
// Tidy shutdown
}
note that this would be resilient to:
queue.put("EOT");
because "EOT" != EOT even though "EOT".equals(EOT) would be true.

What use is there for it? Not much in normal practice but you can always write a class that operates on intern()-ed strings, which can then use == to compare them.
Why it isn't overloaded is a simpler question: because there is no operator overloading in Java. (To mess things up a bit, the + operator IS sort of overloaded for strings, which was done to make string operations slightly less cumbersome. But you can argue that's just syntactic sugar and there certainly is no operator overloading in Java on the bytecode level.)
The lack of an overloaded == operator made the use of the operator much less ambiguous, at least for reference types. (That is, until the point autoboxing/unboxing was introduced, which muddies the waters again, but that's another story.) It also allows you to have classes like IdentityHashMap that will behave the same way for every object you put into it.
Having said all that, the decision to avoid operator overloading (where possible) was a fairly arbitrary design choice.

The == operator compares the reference between two objects. For example, if String x and String y refers to two different things, then the == operator will show false. However, the String.equals() method compares not if they refer to each other, but if the values (ex. "Hello", "World", etc.) are the same.

// A.java
String foo1 = "foo";
// B.java
String bar1 = "foo";
All String literals realized at compile time are added to String Constant Pool. So when you have two different String declarations in two different classes, two String objects will not be created and both foo1 & bar1 refer to the same String instance of value foo. Now that you have same String reference in two different variables, you can just check if those two strings are equal just by using == which is fast because all it does is compare the bit pattern, where as in equals() method, each character is compared and is generally used for two different String instances but same content.
In fact, if you look at equals() implementation in String class, the first check they do is Reference comparison using == because they might seem as different instances to you, but if they're String literals or if they're interned by someone else already, then all you have is a Single reference in two variables.
public boolean equals(Object anObject) {
if (this == anObject) {
return true;
}
// remaining code
}
Also, == is not just for Strings, it's used to compare any two bit patterns, be it primitives or references

1."=="operation of comparison are the values of the two variables are equal, for a reference type variables is expressed by the two variables in the heap memory address is the same, namely the stack have the same content.
2."equals"Whether the two operation variables represent references to the same object in the heap, i.e. whether the contents of the same.

String s = "string1"; creates 1 reference and 1 object in pool String
s1 = "string1"; creates just 1 reference and points to what s is
pointing to.
s == s1 // true
String s2 = new String("string1"); creates 1 object in heap, one in
pool and one reference.
//Since, s2 is pointing to different object so,
s2 == s // false
s1 == s // false
Problem :
So, suppose We want to check, how many unique String object is created and stored in pool by the application while it is running,
We can have a singleton object which can have all the String references stored in an array.
From the previous examples of s, s1 and s2, finally for s and s1, 1 object is created and for s2, 1 object (in total 2).
//If we use equals method, all
s.equals(s1) // gives true
s1.equals(s2) // gives true
//So, number of references present in the array of singleton object will be our
//total number of objects created which equals to 3 // doesn't match actual
//count which is 2
we can use == to check for equality of reference, so if reference is equal, we will not increment our count of unique String object in pool, and for every non equal result, we will increment the count.
here,
for
s // count = 1
s1 == s // count remains same
s2 == s // false, so count = 1 + 1 = 2
//We get total number of unique String objects created and got stored in pool using ==

Simple answer...
Why would one want to compare String references ?
Because they want to compare String values in a very fast way.
Strings are not always interned(). String constants are, but it is possible that the string was created manually on the heap. Using the intern() on a manually created string allows us to to continue using reference comparison on our strings for value comparison.
What is the justification of not overloading it, so that it does a deep comparison ?
Because Java does not have operator overloading as a design decision
Operator '==' is a reference operator always, and equals() is a value method always. In C++ you can change that, but many feel that simply obfuscates the code.

Checking references is Faster compared to checking the entire Strings' equality.
Assume you have Large Strings (URLs or DBMS queries), a have multiple references to them. To check if they are equal, either you can check character by character or you can check if they both refer to the same object.
In fact, equals method in java first checks if the references are same and only if not goes ahead and checks character by character.
Java is full of references and hence, you might need a case where you need to check if two variables are referring to the same String/Object rather than both having each copy of the same String so that you can update string at one place and it reflects in all variables.
To do so, equals method does not help as it checks the copies to be equal as well. you need to check if they both refer to the same object and hence == comes into picture.

It seems that this was asked before and received quite a popular answer here:
Why didn't == operator string value comparison make it to Java?
The simple answer is: consistency
I guess it's just consistency, or "principle of least astonishment".
String is an object, so it would be surprising if was treated
differently than other objects.

Although this is not the fundamental reason, a usage could be to improve performances: before executing a heavy computation, "internalize" your Strings (intern()) and use only == for comparisons.

What I am asking is what uses does the == operator have for String class in Java ?
What is the justification of not overloading it, so that it does a deep comparison ?
== and equals have altogether different uses.
== confirms if there is reference-equality
Equals confirms if the objects contains are same.
Example of reference-equality is IdentityHashMap.
There could be a case in which Only the object inserting something to IdentityHashMap has the right to get/remove the object.
overloading reference-equality can lead to unwanted complexity for java.
for example
if (string)
{
do deep equality
}
else
{
do reference-equality
}
/*****************************************************************/
public class IdentityHashMap extends AbstractMap implements Map, Serializable, Cloneable
This class implements the Map interface with a hash table, using reference-equality in place of object-equality when comparing keys (and values). In other words, in an IdentityHashMap, two keys k1 and k2 are considered equal if and only if (k1==k2). (In normal Map implementations (like HashMap) two keys k1 and k2 are considered equal if and only if (k1==null ? k2==null : k1.equals(k2)).)
This class is not a general-purpose Map implementation! While this class implements the Map interface, it intentionally violates Map's general contract, which mandates the use of the equals method when comparing objects. This class is designed for use only in the rare cases wherein reference-equality semantics are required.

Why is a certain boolean false though it should be true? [duplicate]

This question already has answers here:
How do I compare strings in Java?
(23 answers)
Closed 9 years ago.
I wrote a Java program that should support English and German language. If a parameter is set and if it equals "english" or "English", it shall call a method that does the English version and if there is no parameter or it doesn't equal "English" or "english", it shall call the method for the German version.
However, (args[0]=="english"||args[0]=="English") is false no matter what my parameter is, even if it should be true and I don't get why that's the case.
Here is the main method, the other ones aren't important, so I'll leave them away.
public static void main(String[] args){
boolean input=args.length==1;
System.out.println(input);
boolean mode = false;
if (input) mode=args[0]=="English"||args[0]=="english";
System.out.println(mode);
if(input&&mode) english();
else german();
}
Does anyone have a clue why it won't be true, regardless of my parameter?

Use the equals() method for String value comparison.
args[0].equals("English")||args[0].equals("english")
or even better(in this case)
args[0].equalsIgnoreCase("English")
== is for object reference comparisons. Don't use it for comparing the values.

You can't compare strings in Java in this way because Java Machine compare pointers to the string objects. To make correct compassion use function equals:
if( "english".equalsIgnoreCase( args[0] ) ) {
// English language
}

Strings should not be compared with == but with equals method.
Use
args[0].equalsIgnoreCase("English")
it will compare for both, "english" and "English".
better version is
"English".equalsIgnoreCase(args[0]);
this will make sure, if args[0] is null, i.e no argument in your case, it will not throw NPE.
Explaination : Because in Java, == compares objects not there values, i.e if two references are holding same object or not. Objects content are compared with equals method of Object class, which String class overrides.

Difference in string comparison result b/w == and String#replace with == [duplicate]

This question already has answers here:
Closed 10 years ago.
Possible Duplicate:
String comparison and String interning in Java
I have small doubt regarding String comparisons in Java, consider the following code:
if("String".replace('t','T') == "String".replace('t','T')) {
System.out.println("true");
}
else {
System.out.println("false");
}
The above code always print's false, where as if I try like this:
if("STring" == "STring") {
System.out.println("true");
}
else {
System.out.println("false");
}
It will always print me true. Yes, I know String comparisons should be done with String.equals() or equalsIgnoreCase() method. But this is one of the question was asked in interview and I am confused. Can anyone guide me on this behavior?
As per my knowledge, in code snippet 1, "String.replace('t','T') is returning object, so object comparisons returns in false. Am I right?

"String.replace('t','T') is returning object, so object comparisons
returns in false. Am I right?
Yes, as for this case, you are right. String#replace(or any method of String class for that matter), will return a new String object (You can guess why? Immutability). And thus you would have to do the comparison using equals method, to compare their contents.
Now, in the second case: -
"STring" == "STring"
You are comparing two string literals. Now, since String literals are interned in Java, so both the literals are same (in the sense, they point to the same memory location), and hence == comparison gives you true.
The difference in comparison using == and equals is that, == compares the reference value - i.e value of memory location of objects, which will be different for two different string objects, as you are having in first case. Whereas, equals compares the actual content in those objects.

"String.replace('t','T') is returning object, so object comparisons
returns in false. Am I right?
Yes, == compares object references, and your first code is comparing two different objects.
As far as the second code is concerned its due to string interning.

ok lets do it like this, your both String objects "String" are referering to the same object.
So they are "basicly" equal. That is a thing the compiler does for you
but the method replace, does create and return a new String object, and that is why your second code is not equal.

Java always compares the basic types (int, byte, etc) or references for objects when using ==.
The java compiler optimizes the two string constants you entered to use the same object, thus the same reference, thus the == return true

DO this way
("String".replace('t','T').Tostring() == ("String".replace('t','T')).ToString()
This will solve your problem because the replace statement should be converted to string before eveluation.
You can also user the String.Equals for this or better you use ignore case as you mention in your question.

Try this:
if(string1.equals(string2)){
...
}

Not working String comparison in Java

This code is not working:
String name = "Bob";
String name1 = "Anne";
if name = name1;
System.out.println ("Hello");
I am a beginner in Java, please help we with this code. I am trying to compare two strings.

You want:
if (name.equals(name1))
Note that you don't want
if (name == name1)
which would be syntactically correct, but would compare the two string references for equality, rather than comparing whether the objects involved represent the same sequence of characters
Further, note that even the top version will simply perform an ordinal comparison of the UTF-16 code units in the string. If the two strings logically represent the same characters but are in different forms, you may not get the result you expect. If you want to do a culture-sensitive comparison, you should look at Collator.
Additionally, I'd recommend that if you're really new to Java, you start off with console apps and/or unit tests to explore the language instead of JSP - it'll give you a much smoother cycle while you're learning the basics, and a simpler environment to work in.
Even more additionally, the code given at the top will throw a NullPointerException if name is a null reference. You should consider what you want to happen at that point - if the string being null would represent a bug anyway, then the exception is probably appropriate; otherwise, you might want to change the code. One useful method in Guava (which is chock full of good stuff) is Objects.equal:
if (Objects.equal(name, name1))
The method return true if both arguments are null, false if exactly one argument is null, and the result of calling equals() if they're both non-null. Very handy.

Your Mistakes :
You are using "=" operator to compare strings. It is not a conditional operator, it is an Assignment operator. The conditional operator in Java is "==" which is used to compare two values if they are equal or not. You even cannot use this one for Strings.
You are writing like this :
if name = name1;
System.out.println ("Hello");
You have put a semi-colon at the end of if statement. So it will do nothing (if your condition is supposed to be right, which is not in this case) however the condition is true or not.
You are missing parantheses around the condition given in if statement.
Synatx of if statement is : if(condition)
So it is must to write "()" around your condition.
Now, for comparing Strings, String class gives us methods like :
stringOne.equals(stringTwo)
It checks for exactly the same string.
or
stringOne.equalsIgnoreCase(stringTwo)
It will ignore Caps-Small letter case.

You must compare the two variables like this
if (name.equals(name1))
This should work and not the way you did it!!!

if(name.equals(name1))
System.out.println("Hello");
== works only when you compare primitives like int or long.
If you want to compare String you have to use either equals() or compareTo(). Single = is an assignment not comparison by doing name=name1 you essentially assign string name1 to variable name.

Your posted code isn't really Java. In addition, you don't compare '==' with the assignment operator '='. Finally, to do proper comparison of 'Object's or anything descended from Objects you need to use the .equals(...) method.
Comparing with == means "is the same object", not "is an object with the same value". The difference seems to be small; but, if you opt to compare objects by their value, it is not small at all. Two Objects can be created with identical "values", and only .equals(...) allows you to consider those two Objects to be the same.

Weird Java Behaviour in string comparison [duplicate]

This question already has answers here:
Closed 11 years ago.
Possible Duplicate:
Java string comparison?
I have encounter the following problem, I have an object called "lang", is a result from a method LanguageDetector.detect() which output a string.
lang = LanguageDetector.detect();
So I would want to check whether the language is english, so I am checking,
lang == "en"
The following screen is my debug screen, my lang is showing "en", however my lang == "en" is showing false and lang.toString() == "en" is false, does anyone encounter following problem before and have a possible solution?

Use equals() method of String object instead of direct comparison.
String first = new String("Hello");
String second = new String("Hello");
first == second will return false.
first.equals(second) will return true.

In Java, == always does a reference comparison. You need a value comparison though (with the equals() method for instance).

You're comparing the references to the Strings rather than the contents of the strings themselves. See here for more info.
Note that this issue doesn't apply just to Strings, but to all objects. As such, you may have to define appropriate equals() methods for any objects you create yourself.
Additionally String interning will confuse matters if you're not careful. See here for more details.

Use lang.equals("en") instead of lang == "en". The latter compares the two string references for equality, whereas the former compares the contents of the two strings.
See http://www.devdaily.com/java/edu/qanda/pjqa00001.shtml for an overview of different string comparison methods in Java.

By using == you are checking that both string references point to the same object.
For strings that are created on the fly, and not interned, this will equal false.
To compare the strings for equality, letter by letter, use string1.equals(string2) or even string1.equalsIgnoreCase(string2).

Use "en".equals(lang) instead of lang == "en"

Its better to use the equals as said
but if its necessary for performance reasons you can try
the intern() function.
lang.intern() == "en"