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.
Related
I'm into Java since a short time, and I was wondering: Strings are in fact objects, but I heard that in assigning them a value and retrieving it they act quite differently, almost as if they were primitive types... could someone make it more clear?
What do I exactly have to care about when I declare/edit/access a string compared to other objects?
First of all Java has string literals. That means you may write String foo = "bar";.
String are immutible (once you create one, you can't change it) and it helps JVM to do one trick called "string pool". String literals are stored in pools, and in the following example both foo and bar may point to one instance of string. String foo = "baz"; String bar = "baz". You may even compare them with ==, but you should never do that. How ever, equals() method (which you use to compare strings in Java) may benefit from it since it does not need to compare strings if both vars point to the same string.
Please check this topic for more info What is the Java string pool and how is "s" different from new String("s")?
In java we need to compare objects using .equals() instead of ==. But why can't the compiler do this for us? For example:
if (myString == myOtherString){
doSomething();
}
why cant the compiler go "oh, we're comparing objects!", and change it to this:
if (myString.equals(myOtherString))
Why do we do this manually?
Edit - Guys, I know the difference between == and .equals(). You can stop telling me how ignorant I am. My question was why not just substitute .equals() since its what you want 99% of the time. I have learned that there are cases where knowing if two objects are truly the same reference is useful.
The == operator and equals() often do quite different things. It's only the default implementation of equals() inherited from Object that reverts to using ==. (String is a good example: strings that are equal() are often not ==.) Also, the first example in your code will execute fine if myString is null, while the second will throw an exception.
Sometimes you really just want to know if two object references are to the same object, not whether they refer to objects that have "the same contents" (the meaning of which is usually what equals() implements). Automatically converting == to equals() would be a bad idea.
There is a difference. equals(Object) checks if two objects are equal - i.e., have the same state. The == operator checks if two references indeed point to the same object. It isn't a common usecase, but it definitely has its usages - e.g., to borrow from IdentityHashMap's documentation:
A typical use of this class is topology-preserving object graph transformations, such as serialization or deep-copying. To perform such a transformation, a program must maintain a "node table" that keeps track of all the object references that have already been processed. The node table must not equate distinct objects even if they happen to be equal. Another typical use of this class is to maintain proxy objects. For example, a debugging facility might wish to maintain a proxy object for each object in the program being debugged.
Is there a reason the java compiler cannot just substitute .equals for == when comparing objects?
Java uses both equals() and ==
When you use == to compare objects, you are comparing whether the 2 objects reference the same instance.
When you use .equals(), most of the time you will be comparing one or some of the attributes of the 2 objects. (Comparing the 2 objects' content)
Example for use of == in optimization
public boolean equals(square s){
if(this == s){ //If s and this object is the same instance
return true; //return true straight away, no further checking needed
}
return (this.length == s.getLength() && this.breadth == s.getBreadth());
}
I have a scenario like this -
String s = "abc", t="abc"; //LINE 1
System.out.println(s==t); // definitely it would return true; //LINE 2
s=s+"d"; t=t+"d"; //LINE 3
System.out.println(s==t); // output would be false; but why??
s=s.intern(); t=t.intern();
System.out.println(s==t); // it would return true;
I wanted to know why the second print statement returned false. Please provide me any reference link which explains the same.
While creating t at line 1; intern was called and it pointed to "abc" why not intern was called at line 3?
java strings are immutable.
that means that when you do something like s=s+"d" youre actually creating a whole new string, and assigning it to s.
on top of that, the compiler does constant detection and allocation, so that when you write s="abc", t="abc" the compiler re-uses the same reference and your code is effectively s=t="abc"
so you start with the exact same string instance (thanks to compiler optimization) and turn it into 2 identical yet different strings, at which point s==t is false (s.equals(t) would have been true, as it compares the contents and not the address in memory).
next up is intern(). what intern() does is looks up an identical string in the string cache and returns it. if it doesnt find an identical entry it places the argument provided into the cache and returns the argument. so s=s.intern() places s into the string cache and returns it (so s is unchanged) but the following call t=t.intern() actually returns s, so that s==t again.
Strings are "special" Java objects.
The JVM tries to reuse the same references (that's why String s = "abc", t="abc"; causes s and t to point to the same instance), however, when working on instances (like t=t+"d") a new instance gets created, thus, the references are not the same
In order to compare strings you have to use the .equals() method.
intern() causes to create a canonical representation out of the string pool inside the String class (
http://docs.oracle.com/javase/7/docs/api/java/lang/String.html#intern%28%29)
String s = "abc", t="abc";
s == t is true because Java automatically interns String literals. In this case the String literal "abc" has been interned and both s and t point to that same instance. Hence s == t is true.
s = s + "d"; t = t + "d";
Strings in Java are immutable. Hence what you are assigning to s and t are two new Strings that have been constructed. Therefore they do not point to the same instance. This is why s == t returns false.
s = s.intern(); t = t.intern();
Here you have forcibly interned the string in s.intern(). Since both s and t contain the same string values, the JVM sees that t is the same and makes it point to the same interned-instance as s. Hence s == t is true.
As a general note, establishing the equality of strings should be done via .equals() and not ==; == only compares references for reference-types and not values.
Java Language Specification explicitly covers this particular situation. Here is a quote from chapter 3.10.5. "String Literals":
Moreover, a string literal always refers to the same instance of class String. This is because string literals - or, more generally, strings that are the values of constant expressions (ยง15.28) - are "interned" so as to share unique instances, using the method String.intern.
As you can see, only constant expressions are interned. So, first four lines of your code are equivalent to:
String s = "abc".intern(), t="abc".intern();
System.out.println(s==t);
s=s+"d".intern(); t=t+"d".intern();
System.out.println(s==t);
Expressions s+"d" and t+"d" aren't constant and, thus, aren't interned.
JLS even provides an example with useful notes. Here is the relevant part:
package testPackage;
class Test {
public static void main(String[] args) {
String hello = "Hello", lo = "lo";
System.out.print((hello == ("Hel"+lo)));
}
}
Output: false
Note: Strings computed by concatenation at run time are newly created and therefore distinct.
Because when you concatenate Strings you generate a new object reference except when they are literal Strings.
Note that the intern of both Strings point to the same literal String object reference.
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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)){
...
}
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.