public class HelloWorld { public static void main(String[] args) { SuperBase a1 = new SuperBase(); SuperBase b1 = new Base(); Base c1 = new Derived(); a1.print(new SuperBase()); b1.print(new Base()); c1.print(new Derived()); } } class SuperBase { void print(SuperBase a) { System.out.println("superbase"); } } class Base extends SuperBase { void print(Base a) { System.out.println("base"); } } class Derived extends Base { void ... |
Overloading lets you to reuse the same method name repeatedly, but with varying outputs and parameters. Method overloading is usually associated with but not limited to inheritance. You can overload methods in the same class, and you can overload methods defined in a superclass. You most likely have been calling overloaded methods since you started programming in java. The println() method ... |
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but if i have a constructor of argument int and another constructor of argument long and i pass 2 as a value than it will be received by the constructor with int as an argument and if i pass 2147483650 as a value(which is too large for int),it won't compile (obviously) but when i comment the int-constructor(with int as an argument) ... |
Hi guys, In java when we overload a particular method the return type is made optional i.e. it may or may not be same but the method signature(parameter list) must be different. Why the same return type is made optional in the case of method overloading i.e. one can overload method with same return type or differnect return type. Where there ... |
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I was expecting some kind of exception at runtime from the below code but to my surprise the code ran without any issues and the output was "SubClass". From testing similiar code I understood that it chooses the method that takes argument which appears more lower in the inheritance tree. It should work fine if some super class/sub class reference was ... |
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The second method can never be an entry point into a Java application. Period. But then again why would you need it to? All the arguments you pass at the start of the application are available through the args String[]. If you want to call the second method from your main entry point, you just have to wrap the arguments in ... |
It is the ability of the object to behave differently for the same message. Overriding gives the benifit of having the behaviour specific to the subclass.->Which method to be called is known at compile time.Addng on the access level can be less restrictive then the overridden method.And can throw runtime exception even if the overridden method doesn't. Overloading cleanliness of code ... |
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This is from the Bates and Sierra book. "When a class has overloaded methods, one of the compiler's jobs is to determine which method to use whenever it finds an invocation for the overloaded method. Let's look at an example that doesn't use any new Java 5 features:" class EasyOver { static void go(int x) { System.out.print("int "); } static void ... |
I have following example Given the following class definition, which of the following methods could be legally placed after the comment //Here public class Rid{ public void amethod(int i, String s){} //Here} 1)public void amethod(String s, int i){} 2)public int amethod(int i, String s){} 3)public void amethod(int i, String mystring){} 4) public void Amethod(int i, String s) {} Answer should be ... |
There are a set of rules that determine which overloaded method gets called when there are more than one that could be called, like in this case. They can get a little tricky, but the overriding principle is "the method that most closely matches the argument type is called". So, in this example, 5 is an int. It could be autoboxed ... |
class UseAnimals { public void doStuff(Animal a) { System.out.println("In the Animal version"); } } class Animal extends UseAnimals { public void doStuff(UseAnimals ua) { System.out.println("In the Horse version"); } } class Test { public static void main(String[] args) { UseAnimals ua = new Animal(); UseAnimals use = new UseAnimals(); ua.doStuff(use); } } |
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When you override, you create a method that is only seen by the sub-class. So if you have an A object and call the doStuff() method, you'll get the version defined in the A class. When you have a B class, you'll get the B-class version. Note that due to something called "polymorphism", it doesn't matter what the REFERENCE type is. ... |
Hi, i want to know if overloaded methods in java use early binding or dynamic binding. some say that every method in java is dynamically bound, unless the method is static or final, no matter if the method is overloaded or not. and some say that overloaded methods use early binding. which one is right? |
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Hi Guys, I tried to figure out the below codes but still i can't get the correct answer based on my simulation on paper. The answer is letter D. How these codes derived the answer? Can anybody explain in a simple way. Thanks. Donald 3. class A { } 4. class B extends A { } 5. public class ComingThru { ... |
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Varun Nayudu wrote:Hi, I want to ask this simple question 1>What is the need for overloading method? 2> Why cant we create a different method instead of overloading the method? Well, required is a bit strong but overloading is one the more powerful and useful features of Object Oriented programing. For example if we have a base class of People with ... |
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Nothing now it's all theoretical and there's no default constructor. Thinking about it though I'd try to have something like. Java Code: public *type* mutate(){ //randomly generate DNA sequences if(sequence is Terra based){ return Terra object; }else if (sequence is Aquatic based){ return Aquatic object; }else if (sequence is Avian based){ return Avian object; } } So a program designed where ... |
Yes that's overloaded, assuming those methods are in the same class anyway, or one of them is in a subclass. The not overloaded methods cannot be in the same class (can't have two methods with the same signature in the same class)...so I assume you mean the second one is in a subclass? In which case that's overriding. And finally, a ... |
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Hi, i'm consufed with this... below code class Tazan has overriden and overloaded methods, (it has both) Bt when it is called with reference variable, i don't get the expected output.. My idea WAS, overloading reference type and overridding actual object type.. Please help me to verify this Thanks a lot Output of following prgram is Animal class Animal{ public void ... |
Overriding comes in handy for when you want to change functionality of some method based on a class. For example, the toString method in the object class doesn't print much useful. With overriding you can change it to print something more useful for different types of classes. Overloading is useful to add flexibility to a method. What if you want to ... |
Is suppose my point, if I have to have one beyond "Gee, How DOES that work?... that's Interesting" is beware that every string concetenation using the + operator causes some new objects (don't know one, several?) to be created behind the scenes... which just may have a performance impact... so maybe I'd use + to build an SQL query, but NOT ... |
Cikamani wrote: what is the situation it to be used ? If you require the same operation based on different types of parameters or different number of parameters. If you have f.i. a function that does something based on a number of fields, you could overload that to supply variants that have less parameters, but supply a default value for the ... |
1st i wanted to try method overloading by using the same show method with different no of parameters. It gave an output This Method is empty from 1st method. Should i compulsorily declare the classes abstract. public class MtdLoading1 { void show() { System.out.println("This method is Empty"); } void show(int i,int j) { int k = i+j; System.out.println("This Method has 2 ... |
I. If within the same class definition, if i have 2 overloaded methods i.e. methodA and methodB and both having the exact same signature will this class compile or not? II. If in the above question the return type of methodB is a subtype of methodA, then will this class compile? Remember both have the same method signature. |
Actually, that would be a side effect code style. Side effects with Object Oriented Programming, in general is not a good idea, and is dicouraged. Changing the method name in this case in probably the best way. Although, is there any reason why you can not just have one method return the most accurate value, and cast it to the different ... |
public void maximum(num1,num2,num3){ return maximum(maximum(num1,num2),num3); } protected void addImpl(Component comp, Object constraints, int index) { } } public static void main(String[] args) { new TestOverLoading(); } } }{{code} the error message is:- **H:\My-JAVA Programs\TestOverLoading\src\TestOverLoading.java:34: expected** **public void maximum(num1,num2){** **H:\My-JAVA Programs\TestOverLoading\src\TestOverLoading.java:34: expected** **public void maximum(num1,num2){** **H:\My-JAVA Programs\TestOverLoading\src\TestOverLoading.java:38: ';' expected** **return num2** **H:\My-JAVA Programs\TestOverLoading\src\TestOverLoading.java:41: expected** **public void maximum(num1,num2,num3){** **H:\My-JAVA Programs\TestOverLoading\src\TestOverLoading.java:41: ... |
System.out.printf("Num 1\tNum 2\tResult", sub(7,3), sub(13.5,12.6), sub(9.9, 3), sub(6,2.9)); } public static int sub(int num1, int num2) { return (num1 > num2) ? num1 - num2 : num2 - num1; } public static int sub(double num1, double num2) { return ((num1 > num2) ? num1 - num2 : num2 - num1); } public static int sub(double num1, int num2) { return ((num1 ... |
I am starting to write a program that uses method overloading to compute the average of 2 integers, 3 integers, 2 doubles and 3 doubles, where the command line prompts the user to input numbers.... i am having some trouble finding a place to start my program..... reccomendations would be very helpful. thankyou |
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Hi, Thank you for your precise answer. I just thought that JVM will be clever enough to recognize that although there is a reference type of Object in Objects' equals method, the real object is of type Moja and therefore the one of type Moja method will be executed. So as I said earlier, I'm touching the topic of multiple polymorphism, ... |
Hello, I have a method with string parameters in it and want to use the same method name with different parameter combinations for doing something. So I have overloaded this method: public void myMethod (String first, String second, String third) { ? } public void myMethod (String first, String second) { ? } public void myMethod (String first, String third) { ... |
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(1) the list of argument types (int, int, float) must be in different order like (int, float, int) , (float, int, int) , (int, int), (int, float) ... in order to overload? (2) the list of arguments (d, e, f) can be in any order like (d,e,z) , (a, b),(dx,ey,fz) or anything else? |
Your code doesn't actually use the return value from the roll methods. You should just declare it to return void and then remove the return statements. If you're rolling dice, then you're producing positive integers. Rather than using Math.random, you should probably use java.util.Random.nextInt. Do you need to keep a running total? |
hashdata wrote: What is the real-time usage of these concepts...? Overriding is used when two classes react differently to the same method call. A good example is toString(). For Object it just returns the class name and the identityHashCode, for String it returns the String itself and for a List it (usually) returns a String representation of the content of the ... |
Hi guys, In java when we overload a particaular method the return type is made optional i.e. it may or may not be same but the method signature(parameter list) must be different. Why the same return type is made optional in the case of method overloading i.e. one can overload method with same return type or differnect return type. Where there ... |
Methods aren't defined in packages. They're defined in classes. Does that give you a clue? package testPkg; public class Main { int resI; double resD; void meth(int i,int j) { System.out.println("Result of addition : "+(i+j)); } int meth(int i,int j) { return i+j; } double meth(int i,int j) { double res = i+j; } public static void main(String[] args) { Main ... |
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Ah, i think i have just realised somthing from looking at what they are. When i make the call in my control class, it looks for the method which has LibraryItem in its signature. Therefore, the subclasses methods at this current state are useless. So what i need to do is override the method. So back to what i was asking, ... |
Hey new to java. I am also from morgan. You did it right, but placed variables before the main method. Your method must be: public class Pay { public static void main(String[] args ) { double hours = 37.75; other variables here method call here // You know I am in your class. Good luck for your INSS 250 on Monday ... |
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cosmic_cow wrote: What I don't get is how you can deal with the user's inputs in such a way as to differentiate them. If I declare the array as a double array, even if the user inputs all int values (1, 2, 3, etc), won't it cast them as doubles and call the double function? And if I declare it as ... |
Which method you want to call is something that has to be decided at compile time not runtime. You could use reflection to call the methods instead, but that's a very bad idea unless you have a specific good reason for using it. Your example isn't sufficient to show such a reason. Why do you want to do this? |
Hello all, I am contemplating on a small issue: I have a library that exposes a public method with a list of parameters. It turns out that every once in a while I need to add and/or remove some parameters from the API, but to maintain backwards compatibility, I must not change the existing method's signature, but rather add another one ... |
Hello all, Recently there is a need for using same method with different signatures in an IDL. We tried to do so but it failed. Can anyone please help regarding this? i.e. how to use same method (same name) with different parameters in IDL and map it to java programming? Thanks in advance. |
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Otherwise, if all the maximally specific methods are declared abstract, and the signatures of all of the maximally specific methods have the same erasure ([4.6]), then the most specific method is chosen arbitrarily among the subset of the maximally specific methods that have the most specific return type. However, the most specific method is considered to throw a checked exception if ... |
Assume Dog IS-A Animal class test { public static void main(String[] args){ Animal a = new Dog(); foo(a); } void foo(Dog d){ d.printme(); } void foo(Animal a){ a.printme(); } }//end of class test class Animal{ void printme(){ System.out.println("i am in animal"); } }//end of class animal class Dog{ void printme(){ System.out.println("i am in Dog"); } }//end of class dog --- Question:what ... |
Ok... I have a physics simulation with some different objects... let's say Sphere, Box and Complex (custom 3D-Model) Now, I want to write a class that checks if two Objects collide. wouldnt simply calling CollisionClass.doCollide(object1, object2) be the most elegant version? How qould you design such a class? [Edit] err, no, Multi-Dispatch is not what I want to do. What I ... |
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Dear JK2004, These two threads (this and your other vague one on interfaces) make me think something is very wrong here. I think that either this is someone's else account which you have hi-jacked and if so you should stop and get yourself a new account ASAP. Or you are in fact the same person and I don't even want to ... |
I tired to find a similar post about this but was unable to. I discovered this design issue when working with C#. C# didn't handle like I thought it should, (at least it didn't handle as I was used to Java handling). But the more I thought about it the more I was confused about why Java worked this way. consider ... |
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it is valid. You are extending the method, not overriding it. As such, get rid of the Override statement and all will be well. My thinking goes, that you are not changing the original method one bit. It is still present and functional unchanged in your new class. Thus you have not overridden it at all and you should not state ... |
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Let's talk method headers: If you get stuck on setting up your methods...here's a hint. All your methods will have the same name. The number of parameters will be different. They will all calculate the avereage of the numbers they receive in the method call...so be sure to use the correct number of arguments in your method calls! An example of ... |
hi friends, i want to know that why we have to go for overloading ,instead the same task can be done using another method having different name. Please reply a perfect answer . i know that using same mehod name we are trying to work with different parameters, i want to know more specific reason for method overloading |
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Kayaman wrote: They're bonus knowledge. Fix your attitude, bub. I don't think there is anything wrong with my attitude. I just don't take them as "bonus knowledge" because you haven't given any explanation of the meanings. Therefore, I took them as a humiliation instead at the first place. If you would like to give out extra information for something else on ... |
Overloading: Overloading gives us the power to call a single method and pass it different parameters and the compiler at runtime decided which suitable method to call. Take an example, Let us consider two functions area(float a); area(int a ,int b); for calculating area of circle we can call like area(2); for calculating area of rectangle we can call like area(4,2); ... |
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