This question already has answers here:
Operator overloading in Java
(10 answers)
How does the String class override the + operator?
(7 answers)
Closed 6 years ago.
Everywhere on the internet i found that java doesn't support Operator overloading but i am confused somehow.Because if that is so then how is the "+" operator able to add both constants and strings?
Any explanation would be appreciated
A language is said to support operator overloading when you can overload operators, i.e. make them do something that is not built into the language. Not when the langauge uses the same operator for two different things.
You can use these operators because it was implemented in the language how they behave. But there is no support for operator overloading in java.
You can create methods such as add(Object o) which is basically the same as what operators do except that it doesn't look as good.
Some languages, such as C#, allow you to overload the operators. In other words, you can define what is meant by == or ++, etc. This can be very useful in scenarios where you wish to use == to test for equality of objects instead of using a .equals() method, for instance.
Here's a tutorial for C#, which shows you how to do it with a + operator:
http://www.tutorialspoint.com/csharp/csharp_operator_overloading.htm
Or this:
https://msdn.microsoft.com/en-us/library/aa288467(v=vs.71).aspx
And you can't do this in Java at present.
Java is a Strong Static Casting so does that mean there is no use for "==="
I have looked at tons of documentation and have not seen Identical Comparison Operator.
=== is useful in weak typed languages, such as Javascript, because it verifies that the objects being compared are of the same type and avoids implicit conversions.
=== has absolutely no use in a strongly typed language such as Java because you can't compare variables of different types without writing a specific method for doing this.
For example, if you want to compare an int to a String in Java, you will have to write some special method as such:
boolean compareIntString(int i, String s) {
return (i == parseInt(s));
}
But this is pretty much overkill. (And as you'll notice, as written, this method only accepts an int and a String. It doesn't accept just any two variables. You know before you call it that the datatypes are different.)
The main point is, that while you can do i == s in Javascript, you can't do i == s in Java, so you don't need ===.
I guess, the short answer is that Java's == is Javascript's ===. If you want to emulate Javascript's == and compare two items, ignoring data type, you'll have to write a custom method which accepts generic data types as arguments... and figure out the logic on comparing, at a minimum, all the possible combinations of Java's primitive data types...
No java does not have === operator. Reason is pretty well explained by nhgrif. Here is the list of operators in java and their precedence:
Source: http://docs.oracle.com/javase/tutorial/java/nutsandbolts/operators.html
I am attempting to create a unsigned integer class.
public class UnsignedInteger extends Number implements Comparable<UnsignedInteger>
{
...
}
Is there a way to implement operators such as; +, -, *, /, <<, >>, |, ^, >>>, <<
Java does not support Operator Overloading. The only option you have is define methods like add(), subtract(), multiply(), etc, and write the logic there, and invoke them for particular operation.
You can have a look at BigInteger class to get an idea of how you can define methods to support various operations. And if interested, you can even go through the source code, that you can find in the src folder of your jdk home directory.
There are already 5 answers saying that you cannot overload operators, but I want to point out that you can not use arithmetical operators on objects at all. They only work with primitive types (int, double, etc).
The only reason the following code compiles
Integer a = 1, b = 2;
Integer c = a + b;
is because the Java compiler compiles it as
Integer a = Integer.valueOf(1), b = Integer.valueOf(2);
Integer c = Integer.valueOf(a.intValue() + b.intValue());
If you want this to work for your UnsignedInteger, you have to extend the javac (it is possible, though).
No you cannot override operators in Java.
It's not possible to override operators in Java. What you can do is define methods to represent the operations, like BigDecimal or BigInteger in the standard library do.
There is a javac-plugin (an annotation processor like Lombok) called "Java-OO", which adds operator overloading to Java.
It allows you to add operator overloading to your own classes very easily. In addition to this, many of the built-in classes of the Java API also supports operator overloading when using this plugin.
(For example: Instead of list.get(6) or map.get("hello") you can do list[6] and map["hello"])
All you need to do is to include the .jar on the classpath when compiling with javac.
There are plugins for all major IDEs: Eclipse, Netbeans and IntelliJ IDEA.
No. Java does not support operator overloading.
Java doesnt support operator overloading, they consider it a bad practice, knowing that they overloaded + and += operators for the String class
I program in Java and have been trying to understand exactly what operator overloading is. I'm still a bit puzzled.
An operator can take on different meanings depending on which class uses it? I've read that it is "Name Polymorphism".
Java apparently does not support it and there seems to be a lot of controversy around this. Should I worry about this?
As a last question, in an assignment the teacher has stated that the assignment uses operator overloading, he is a C++ programmer mainly but we are allowed to write the assignment in Java. since Java does not support overloading, is there something I should be wary of?
Operator overloading basically means to use the same operator for different data types. And get different but similar behaviour because of this.
Java indeed doesn't support this but any situation where something like this could be useful, you can easily work around it in Java.
The only overloaded operator in Java is the arithmetic + operator. When used with numbers (int, long, double etc.), it adds them, just as you would expect. When used with String objects, it concatenates them. For example:
String a = "This is ";
String b = " a String";
String c = a + b;
System.out.print (c);
This would print the following on the screen: This is a String.
This is the only situation in Java in which you can talk about operator overloading.
Regarding your assignment: if the requirement is to do something that involves operator overloading, you can't do this in Java. Ask your teacher exactly what language you are allowed to use for this particular assignment. You will most likely need to do it in C++.
PS: In case of Integer, Long, Double etc. objects, it would also work because of unboxing.
Java doesn't allow overloading operators. It uses a very limited kind of operator overloading though, since + does addition or concatenation depending on the context.
If your assignment asks you to implement something by overloading operators, you won't be able to do it in Java. Maybe you should ask the teacher why he allows Java for such an assignment.
If your assignment only asks you to use an overloaded operator, then having your program use + for concatenation and addition would fit the bill. But I would ask the teacher, because I doubt that it's what he expects.
Java apparently does not support it and there seems to be a lot of
controversy around this.
There is no controversy about this. Some people might disagree with the decision, but James Gosling and others decided from day one to leave operator overloading by class developers out of the language. It's not likely to change.
As pointed out by others here, they reserved the right for the JVM to overload operators on a limited basis. The point is that you can't do it when you're developing your own classes.
They did it because there were examples of C++ developers abusing the capability (e.g. overloading the dot operator.)
Should I worry about this?
No. Java won't explode. You just won't be able to do it for your classes. If you feel like you need to, you'll just have to write C++ or some other language.
As to your query about the difference between operator overloading and polymorphism. Polymorphism is a standard OOP concept where an instance of a class may exhibit different characteristics depending on the underlying type. For example in C++:
class Shape {
...
virtual void draw()=0;
}
class Circle :public Shape {
virtual void draw() {...draw a circle...}
}
class Square:public Shape {
virtual void draw() {...draw a square...}
}
..
Shape *s = new Circle();
s->draw(); // calls Circle::draw();
s=new Square(); // calls Square::draw();
Hence s is exhibiting polymorphism.
This is different from operator overloading but you already have been explained what that is in the other answers.
Can either use natural a != b (a is not equal to b) or a.Equals(b)
b.set(1, 0);
a = b;
b.set(2, 0);
assert( !a.Equals(b) );
But java has a limited set of operator overloading than other languages http://en.wikipedia.org/wiki/Operator_overloading
Closed. This question is opinion-based. It is not currently accepting answers.
Want to improve this question? Update the question so it can be answered with facts and citations by editing this post.
Closed 6 years ago.
Improve this question
Why doesn't Java need operator overloading? Is there any way it can be supported in Java?
Java only allows arithmetic operations on elementary numeric types. It's a mixed blessing, because although it's convenient to define operators on other types (like complex numbers, vectors etc), there are always implementation-dependent idiosyncrasies. So operators don't always do what you expect them to do. By avoiding operator overloading, it's more transparent which function is called when. A wise design move in some people's eyes.
Java doesn't "need" operator overloading, because no language needs it.
a + b is just "syntactic sugar" for a.Add(b) (actually, some would argue that a.Add(b) is just syntactic sugar for Add(a,b))
This related question might help. In short, operator overloading was intentionally avoided when Java was designed because of issues with overloading in C++.
Scala, a newer JVM language, has a syntax that allows method overloading that functions very much like operator overloading, without the limitations of C++ operator overloading. In Scala, it's possible to define a method named +, for example. It's also possible to omit the . operator and parentheses in method calls:
case class A(value: Int) {
def +(other: A) = new A(value + other.value)
}
scala> new A(1) + new A(3)
res0: A = A(4)
No language needs operator overloading. Some believe that Java would benefit from adding it, but its omission has been publicized as a benefit for so long that adding it is almost certainly politically unacceptable (and it's only since the Oracle buyout that I'd even include the "almost").
The counterpoint generally consists of postulating some meaningless (or even counterintuitive) overload, such as adding together two employees or overloading '+' to do division. While operator overloading in such languages as C++ would allow this, lack of operator overloading in Java does little to prevent or even mitigate the problem. someEmployee.Add(anotherEmployee) is no improvement over someEmployee + anotherEmployee. Likewise, if myLargeInteger.Add(anotherLargeInteger) actually does division instead of addition. At least to me, this line of argument appears thoroughly unconvincing at best.
There is, however, another respect in which omitting operator overloading does (almost certainly) have a real benefit. Its omission keeps the language easier to process, which makes it much easier (and quicker) to develop tools that process the language. Just for an obvious example, refactoring tools for Java are much more numerous and comprehensive than for C++. I doubt that this can or should be credited specifically and solely to support for operator overloading in C++ and its omission in Java. Nonetheless, the general attitude of keeping Java simple (including omission of operator overloading) is undoubtedly a major contributing factor.
The possibility of simplifying parsing by requiring spaces between identifiers and operators (e.g., a+b prohibited, but a + b allowed) has been raised. At least in my opinion, this is unlikely to make any real difference in most cases. The reason is fairly simple: at least in a typical compiler, the parser is preceded by a lexer. The lexer extracts tokens from the input stream and feeds them to the parser. With such a structure, the parser wouldn't see any difference at all between the a+b and a + b. Either way, it would receive exactly three tokens: identifer, +, and identifier.
Requiring the spaces might simplify the lexer a tiny bit--but to the extent it did, it would be completely independent of operator overloading, at least assuming the operator overloading was done like it is in C++, where only existing tokens are used1.
So, if that's not the problem, what is? The problem with operator overloading is that you can't hard-code a parser to know the meaning of an operator. With Java, for some given a = b + c, there are exactly two possibilities: a, b and c are each chosen from a small, limited set of types, and the meaning of that + is baked into the language, or else you have an error. So, a tool that needs to look at b + c and make sense of it can do a very minimal parse to assure that b and c are of types that can be added. If they are, it knows what the addition means, what kind of result it produces, and so on. If they are't, it can underline it in red squiggles (or whatever) to indicate an error.
For C++, things are quite different. For an expression like a = b + c;, b and c could be of almost entirely arbitrary types. The + could be implemented as a member function of b's type, or it could be a free function. In some cases, we might have a number of operator overloads (some of which could be templates) that could carry out that operation, so we need to do overload resolution to determine which one the compiler would actually select based on the types of the parameters (and if some of them are templates, the overload resolution rules get even more complex).
That lets us determine the type of the result from b + c. From there we basically repeat the whole process again to figure out what (if any) overload is used to assign that result to a. It might be built-in, or it might be another operator overload, and there might be multiple possible overloads that could do the job, so we have to do overload resolution again to figure out the right operator to use here.
In short, just figuring out what a = b + c; means in C++ requires nearly an entire compiler front-end. We can do the same in Java with a much smaller subset of a compiler2
I suppose things could be somewhat different if you allowed operator overloading like, for example, ML does, where a more or less arbitrary token can be designated as an operator, and that operator can be given a more or less arbitrary associativity and/or precedence. I believe ML handles this entirely in parsing, not lexing, but if you took this basic concept enough further, I can believe it might start to affect lexing, not just parsing.
Not to mention that most Java tools will use the JDK, which has a complete Java compiler built into the JVM, so tools can normally do most such analysis without dealing directly with parsing and such at all.
java-oo compiler plugin can add Operator Overloading support in Java.
It's not that java doesn't "need" operator overloading, it's just a choice made by its creators who wanted to keep the language more simple.
Java does not support operator overloading by programmers. This is not the same as stating that Java does not need operator overloading.
Operator overloading is syntactic sugar to express an operation using (arithmetic) symbols. For obvious reasons, the designers of the Java programming language chose to omit support for operator overloading in the language. This declaration can be found in the Java Language Environment whitepaper:
There are no means provided by which
programmers can overload the standard
arithmetic operators. Once again, the
effects of operator overloading can be
just as easily achieved by declaring a
class, appropriate instance variables,
and appropriate methods to manipulate
those variables. Eliminating operator
overloading leads to great
simplification of code.
In my personal opinion, that is a wise decision. Consider the following piece of code:
String b = "b";
String c = "c";
String a = b + c;
Now, it is fairly evident that b and c are concatenated to yield a. But when one consider the following snippet written using a hypothetical language that supports operator overloading, it is fairly evident that using operator overloading does not make for readable code.
Person b = new Person("B");
Person c = new Person("C");
Person a = b + c;
In order to understand the result of the above operation, one must view the implementation of the overloaded addition operator for the Person class. Surely, that makes for a tedious debugging session, and the code is better implemented as:
Person b = new Person("B");
Person c = new Person("C");
Person a = b.copyAttributesFrom(c);
OK Well... we have a very discussed and common issue. Today, in software industry, there are, mainly, two different types of languages:
Low level languages
High level languages
This distinction was useful about 10 years before now, the situation, at present, is a bit different.
Today we talk about business-ready applications.
Business models are some particular models where programs need to meet many requirements. They are so complex and so strict that coding an application with a language like c or c++ would be very time-spending. For this reason hybrid languages where invented.
We commonly know two types of languages:
Compiled
Interpreted
Well, today there is another one:
Compiled/Interpreted: in one word: MANAGED.
Managed languages are languages that are compiled in order to produce another code, different from the original one, but much more complex to handle. This INTERMEDIATE LANGUAGE is then INTERPETED by a program that runs the final program.
It is the common dynamics we came knowing from Java... It is a winning approach for business-ready applications.
Well, now going to your question...
Operator overloading is a matter that concerns also multiple inheritance and other advanced characteristics of low level languages.
Java, as well as C#, Python and so on, is a managed language, made to be easy to write and useful for building complex applications in very few time.
If we included operator overloading in Java, the language would become more complex and difficult to handle.
If you program in C++ you sure understand that operator overloading is a very very very delicate matter because it can lead to very complex situations and sometimes compiler might refuse to compile because of conflicts and so on... Introducing operator overloading is to be done carefully. IT IS POWERFUL, but we pay this power with an incredibly big load of problems to handle.
OKOK IT IS TRUE, you might tell me: "HEY, But C# uses operator overloading... What the hell are you telling me? why c# supports them and Java not?".
Well, this is the answer. C#, yes, implements operator overloading, but it is not like C++. There are many operator that cannot be overloaded in c# like "new" or many others that you can overload in c++... So C# supports operator overloading, but in a much lower level than c++ or other languages that fully supports it. But this is not a good answer to the earlier question...
The real answer is that C# is more complex than Java. This is a pro but also a con. It is a matter of deciding where to place the language: high level, higher level, very high level?
Well, Java does not support op overloading because it wants to be fast and easy to manage and use. When introducing op overloading, a language must also carry a large amount of problems caused by this new functionality.
It is exactly like questioning: "Why does Java not support multiple inheritance?"
Because it is tremendously complex to manage. Think about it... IT WOULD BE IMPOSSIBLE for a managed language to support multiple inheritance... No common class tree, no object class as a common base class for all classes, no possibility of upcasting (safely) and many problems to handle, manage, foresee, keep in count...
Java wants to be simple.
Even if I believe that future implementations of this language will result in supporting op overloading, you will see that the overloading dynamics will involve a fewer set of all the possibilities you have about overloading in C++.
Many others, here, also told you that overloading is useless.
Well I belong to those ones who think this is not true.
Well, if you think this way (op overloading is useless), then also many other features of managed languages are useless too. Think about interfaces, classes and so on, you really do not need them. You can use abstract classes for interface implementations... Let's look at c#... so many sugar syntax, LINQ and so on, they are not really necessary, BUT THEY FASTEN YOUR WORK...
Well, in managed languages everything that fasten a development process is welcome and does not imply uselessness. If you think that such features are not useful than the entire language itself would be useless and we all would come back programming complex applications in c++, ada, etc. The added value of managed languages is to be measured right on this elements.
Op overloading is a very useful feature, it could be implemented in languages like Java, and this would change the language structure and purposes, it would be a good thing but a bad thing too, just a matter of tastes.
But today, Java is simpler than C# even for this reason, because Java does not supports op overloading.
I know, maybe I was a little long, but hope it helps. Bye
Check Java Features Removed from C and C++ p 2.2.7 No More Operator Overloading.
There are no means provided by which
programmers can overload the standard
arithmetic operators. Once again, the
effects of operator overloading can be
just as easily achieved by declaring a
class, appropriate instance variables,
and appropriate methods to manipulate
those variables. Eliminating operator
overloading leads to great
simplification of code.
Java doesn't support operator overloading (one reference is the Wikipedia Operator Overloading page). This was a design decision by Java's creators to avoid perceived problems seen with operator overloading in other languages (especially C++).