I wonder if there is a way to make an own conversion to integer in java. I mean a solution that is comparable to the implementation of the string conversion (toString). I want my class to be interpreted as integer in equation without calling a special function.
class MyClass(){
private int number;
MyClass(int n){
this.number = n;
}
public int toInteger(){
return number
}
}
usage:
MyClass a = new MyClass(2);
int result = 1+a;
result would be 3.
Java doesn't allow operator overdloading/overriding. You cannot do this.
You're describing operator overloading, and it's not possible in Java. The closest thing would be subclassing Number, but + doesn't work with it. For Strings + works because it has been built in as a special case in the language. There's no way to extend + to work with anything else.
Of course, with your class, int result = 1 + a.toInteger(); works. Just a little extra work.
Java is not C++ you can't overload operators like '+' in Java
This cannot be done in java. If you are curious enough, check this post
Why doesn't Java offer operator overloading? which excplains why it cannot be done and compares between java and c++ approaches
Related
I can't find any information on this anywhere and was wondering whether such a use of a class is considered bad practise or not.
Let me explain. I have a class ToDecimalConverter which converts an integer from any base to decimal. However, I now need to add the functionality to also convert fractions. As such, I abstracted the integer conversion into a separate class and created a new class with the purpose of converting fractions. (Code isn't finished so I just added some comments to explain)
public class ToDecimalConverter {
private IntegerToDecimalConverter integerConverter;
private DoubleToDecimalConverter doubleConverter;
public double convert(String number, int baseNumber) {
this.integerConverter = new IntegerToDecimalConverter();
this.doubleConverter = new DoubleToDecimalConverter();
number = this.removeZerosAtBeginningOfNumber(number);
// split the number into integer and fraction so they can be used below:
int decimalInt = this.integerConverter.convert(integerNumber, baseNumber);
double decimalDouble = this.doubleConverter.convert(fractioNumber, baseNumber);
// add them together and return them
}
}
Now, except for the methods that remove the zero's from the start of a number and the method that splits the number into integer and fraction (both of which can easily be abstracted into their own class), the ToDecimalConverter class does nothing but group the integer and fraction converters together.
When searching online, I don't see a lot of classes being used like this. Should this be avoided or not? and if so, what are alternatives?
This meant as a more general question, the above is just to explain what I mean.
Edit: Or should I see it as a sort of mini GoF Facade pattern?
There is nothing wrong with it by default, but I would guess that you could achieve the same result with two methods. something like:
public int convertFromInt(String number, int baseNumber) {
int theConvertedInt = 0;
//Really cool convertion
return theConvertedInt;
}
public double convertFromFraction(String number, int baseNumber) {
double theConvertedInt = 0;
//Really cool convertion
return theConvertedInt;
}
Also, keep in mind that a lot of this conversions are already done by Java native classes like BigInteger, BigDecimal, Integer, Decimal, Double, the Math package and so on.
Not going into the specifics of what your class is doing, there indeed value in grouping several or many function/classes together to from a single unified API.
This is called the Facade design pattern.
The intent is that instead of relying on your client to have to know of the various classes/objects you use internally to achieve a feature and to have to look all over the place inside your implementation code is that you put in place a single entry point for given feature/set of feature. It is much better for discoverability & documentation.
Also this way, you ensure to only provide the public API that is only one or a few classes that make the facade while the implementation remains hidden and can change at any time.
I have been looking for a transparent way to cast an object to a primitive type, something like:
class Example {
double number;
public Example(double number) {
this.number = number;
}
// this or something similar to this
public toDouble() {
return number;
}
...
}
Example ex = new Example(18.0);
double number = ex;
After a few searches i am almost sure that it does not exists, but if there would be a big help.
Update 1
Excuse me if the questions was not very clear.
I am developing a code editor for an internal tool: the idea is write a program with a reduced set of java instructions, translate to java and compile.
I want avoid that the editor have to make a lot of times the same casts or call same calls, and the exprexions sometimes will be very complex to translate.
A better example could be:
the simplified code
Example ex = new Example(18.0);
Example ex2 = new Example(23.0);
Example mean = Mean(ex + 1.0, ex2, 3.0);
(i would prefer avoid Example mean = Mean(ex.toDouble() + 1.0, ex2.toDouble(), 3.0);)
and the function in java
double Mean(double... numbers)
All the primitive types in java have their wrappers: Integer, Double etc.
According to Java tutorials:
Converting an object of a wrapper type (Integer) to its corresponding primitive (int) value is called unboxing.
And this is done pretty automatically, so I am not sure which object else would you like to "cast" to primitive? You can use a method like your toDouble(), which will return primitive, but it won't be casting at all.
You can use the wrapper of double, Double.
Double d = new Double(10.3);
double d1 = d; // This will be auto(un)boxing and not casting
transparent way to cast an object to a primitive type
double number = ex; this is illegal as ex is of type Example and number is of the primitive type double. This really doesn't make sense.
You either need to use toDouble(), or completely remove the Example class and use the wrapper class Double as mentioned above.
There is no analogy to C#`s implicit modifier. In general there are less syntax sugar in Java then in .NET. I dont this this is a problem. You can look inside sources of Integer, Double, Float etc. All they have methods like floatValue(), longValue() and so on. Making casts explicit improves code readability. While implicit casts, like available in C#, are very hard to understand. You can make your own method yourTypeValue(). This is a sort of Java convention.
You cannot cast, and little confusing what you are trying to do.
If you need double then , call toDouble()
double number = ex.toDouble();
or even, as some one commented, use the Double class constructor Double(double value)
Double wrapper = new Double(18.0); //class
double primitiveDouble =wrapper; // primitive
In .NET when I had a "value" that could exist as multiple types I could easily use a TypeConverter for switching between those types (currency types, xml data vs object representation, ect). In Java, I am not sure what the preferred way to handle this situation is. Is there a TypeConverter equivalent in Java?
For someone from a .NET world it is going to be a surprise that there isn't one out of the box. This is because we have things like primitives (int, long) etc, their primitive-wrappers (Integer, Long etc), autoboxing from int to Integer when you require (this is from JDK 1.5).
So we the poor Java developers manually convert things (some examples given above by #Boolean)
Also the endless hassles of using == operators when doing these. For ex: Autoboxed integers of upto 127 are cached.
public void testsimple() {
Integer a = 124, b = 124, c = 500, d= 500;
System.out.println(a == b); //prints true in JDK 1.6
System.out.println(c == d); //prints false in JDK 1.6
}
If you are writing a huge app that needs too much data conversion you can write something on your own. Spring's "TypeConverter" interface can be a decent start.
Use this link http://imagejdocu.tudor.lu/doku.php?id=howto:java:how_to_convert_data_type_x_into_type_y_in_java if you have some trouble
Well you can typecast things... for example if you have a List object that contained strings.. you would grab a String from the list like this:
List aList = new List();
aList.add("Hello World);
String s = (String) aList.get(0);
If you are trying to convert a string to a number.. you would do something similar to this:
String aString = "200";
int i = Integer.Parse(aString);
I cheat when converting an integer to a string by doing this:
int i = 200;
String s = "" + i;
TypeConverter where added to .net partly to allow interactive UI designers to show values as string and let the user edit propriety values of types the UI designer does not understand.
I think TypeConverter is also used by data binding.
The fact you can use TypeConverter in your own software to convert between types you know at compile time is a side effect and not the primary use case they were created for.
As Java has never tried to support “RAD” tools, it has not had the same need for TypeConverter and PropertyDescriptors etc. To some extent .net was designed to allow the same sort of UI development that VB6 enabled.
Please can you tell me if it is possible to overload operators in Java? If it is used anywhere in Java could you please tell me about it.
No, Java doesn't support user-defined operator overloading. The only aspect of Java which comes close to "custom" operator overloading is the handling of + for strings, which either results in compile-time concatenation of constants or execution-time concatenation using StringBuilder/StringBuffer. You can't define your own operators which act in the same way though.
For a Java-like (and JVM-based) language which does support operator overloading, you could look at Kotlin or Groovy. Alternatively, you might find luck with a Java compiler plugin solution.
Operator overloading is used in Java for the concatenation of the String type:
String concat = "one" + "two";
However, you cannot define your own operator overloads.
In addition to all the people pointing out that + is overloaded for Strings, - is also overloaded for both floating point and integer operations, as are * and /.
[edit]
% is also overloaded for floating point, which can be a bit of a surprise for those with a C or C++ background.
Java does not allow operator overloading. The preferred approach is to define a method on your class to perform the action: a.add(b) instead of a + b. You can see a summary of the other bits Java left out from C like languages here: Features Removed from C and C++
As many others have answered: Java doesn't support user-defined operator overloading.
Maybe this is off-topic, but I want to comment on some things I read in some answers.
About readability.
Compare:
c = a + b
c = a.add(b)
Look again!
Which one is more readable?
A programming language that allows the creation of user-defined types, should allow them to act in the same way as the built-in types (or primitive types).
So Java breaks a fundamental principle of Generic Programming:
We should be able to interchange objects of built-in types with objects of user-defined types.
(You may be wondering: "Did he say 'objects of built-in'?". Yes, see here.)
About String concatenation:
Mathematicians use the symbol + for commutative operations on sets.
So we can be sure that a + b = b + a.
String concatenation (in most programming languages) doesn't respect this common mathematical notation.
a := "hello";
b := "world";
c := (a + b = b + a);
or in Java:
String a = "hello";
String b = "world";
boolean c = (a + b).equals(b + a);
Extra:
Notice how in Java equality and identity are confused.
The == (equality) symbol means:
a. Equality for primitive types.
b. Identity-check for user-defined types, therefore, we are forced to use the function equals() for equality.
But... What has this to do with operator overloading?
If the language allows the operator overloading the user could give the proper meaning to the equality operator.
You can't do this yourself since Java doesn't permit operator overloading.
With one exception, however. + and += are overloaded for String objects.
One can try Java Operator Overloading. It has its own limitations, but it worth trying if you really want to use operator overloading.
Just use Xtend along with your Java code. It supports Operator Overloading:
package com.example;
#SuppressWarnings("all")
public class Test {
protected int wrapped;
public Test(final int value) {
this.wrapped = value;
}
public int operator_plus(final Test e2) {
return (this.wrapped + e2.wrapped);
}
}
package com.example
class Test2 {
new() {
val t1 = new Test(3)
val t2 = new Test(5)
val t3 = t1 + t2
}
}
On the official website, there is a list of the methods to implement for each operator !
Or, you can make Java Groovy and just overload these functions to achieve what you want
//plus() => for the + operator
//multiply() => for the * operator
//leftShift() = for the << operator
// ... and so on ...
class Fish {
def leftShift(Fish fish) {
print "You just << (left shifted) some fish "
}
}
def fish = new Fish()
def fish2 = new Fish()
fish << fish2
Who doesnt want to be/use groovy? :D
No you cannot use the compiled groovy JARs in Java the same way. It still is a compiler error for Java.
Unlike C++, Java does not support user defined operator overloading. The overloading is done internally in java.
We can take +(plus) for example:
int a = 2 + 4;
string = "hello" + "world";
Here, plus adds two integer numbers and concatenates two strings. So we can say that Java supports internal operator overloading but not user defined.
Please can you tell me if it is possible to overload operators in Java? If it is used anywhere in Java could you please tell me about it.
No, Java doesn't support user-defined operator overloading. The only aspect of Java which comes close to "custom" operator overloading is the handling of + for strings, which either results in compile-time concatenation of constants or execution-time concatenation using StringBuilder/StringBuffer. You can't define your own operators which act in the same way though.
For a Java-like (and JVM-based) language which does support operator overloading, you could look at Kotlin or Groovy. Alternatively, you might find luck with a Java compiler plugin solution.
Operator overloading is used in Java for the concatenation of the String type:
String concat = "one" + "two";
However, you cannot define your own operator overloads.
In addition to all the people pointing out that + is overloaded for Strings, - is also overloaded for both floating point and integer operations, as are * and /.
[edit]
% is also overloaded for floating point, which can be a bit of a surprise for those with a C or C++ background.
Java does not allow operator overloading. The preferred approach is to define a method on your class to perform the action: a.add(b) instead of a + b. You can see a summary of the other bits Java left out from C like languages here: Features Removed from C and C++
As many others have answered: Java doesn't support user-defined operator overloading.
Maybe this is off-topic, but I want to comment on some things I read in some answers.
About readability.
Compare:
c = a + b
c = a.add(b)
Look again!
Which one is more readable?
A programming language that allows the creation of user-defined types, should allow them to act in the same way as the built-in types (or primitive types).
So Java breaks a fundamental principle of Generic Programming:
We should be able to interchange objects of built-in types with objects of user-defined types.
(You may be wondering: "Did he say 'objects of built-in'?". Yes, see here.)
About String concatenation:
Mathematicians use the symbol + for commutative operations on sets.
So we can be sure that a + b = b + a.
String concatenation (in most programming languages) doesn't respect this common mathematical notation.
a := "hello";
b := "world";
c := (a + b = b + a);
or in Java:
String a = "hello";
String b = "world";
boolean c = (a + b).equals(b + a);
Extra:
Notice how in Java equality and identity are confused.
The == (equality) symbol means:
a. Equality for primitive types.
b. Identity-check for user-defined types, therefore, we are forced to use the function equals() for equality.
But... What has this to do with operator overloading?
If the language allows the operator overloading the user could give the proper meaning to the equality operator.
You can't do this yourself since Java doesn't permit operator overloading.
With one exception, however. + and += are overloaded for String objects.
One can try Java Operator Overloading. It has its own limitations, but it worth trying if you really want to use operator overloading.
Just use Xtend along with your Java code. It supports Operator Overloading:
package com.example;
#SuppressWarnings("all")
public class Test {
protected int wrapped;
public Test(final int value) {
this.wrapped = value;
}
public int operator_plus(final Test e2) {
return (this.wrapped + e2.wrapped);
}
}
package com.example
class Test2 {
new() {
val t1 = new Test(3)
val t2 = new Test(5)
val t3 = t1 + t2
}
}
On the official website, there is a list of the methods to implement for each operator !
Or, you can make Java Groovy and just overload these functions to achieve what you want
//plus() => for the + operator
//multiply() => for the * operator
//leftShift() = for the << operator
// ... and so on ...
class Fish {
def leftShift(Fish fish) {
print "You just << (left shifted) some fish "
}
}
def fish = new Fish()
def fish2 = new Fish()
fish << fish2
Who doesnt want to be/use groovy? :D
No you cannot use the compiled groovy JARs in Java the same way. It still is a compiler error for Java.
Unlike C++, Java does not support user defined operator overloading. The overloading is done internally in java.
We can take +(plus) for example:
int a = 2 + 4;
string = "hello" + "world";
Here, plus adds two integer numbers and concatenates two strings. So we can say that Java supports internal operator overloading but not user defined.