class Overload{
public static void main(String args[]) {
int[] number={1,2,3,4,5,6,7,8,9,10};
int [] num={1,2,3,4,5};
int i;
int sum=0;
sum = f(number);
int sum1= f(num);
System.out.println("The sum is" +sum + ".");
System.out.println("The sum is" +sum1 + ".");
}
public static int f(int[] value) {
int i, total = 0;
for(i=0; i<10; i++) {
total = total + value[ i ];
}
return (total);
}
public static int f(int... x) {
int i, total = 0;
for(i=0; i<10; i++) {
total = total + x[ i ];
}
return (total);
}
}
While compiling the above program I'm getting the error as
C:\Program Files\Java\jdk1.7.0_09\bin>javac Overload.java
Overload.java:30: error: cannot declare both f(int...) and f(int[]) in Overload
public static int f(int... x)
public static int f(int... x)
is nothing but: -
public static int f(int[] x)
with only difference that it does not necessarily needs an argument to be passed. And when you pass individual elements, they are converted to an array internally. So, you are actually passing an array only.
Whereas the later one needs an argument. An empty array at the least.
And both the methods are eligible to be invoked, if you are passing an array as argument.
So the call:
f(new int[] {1, 2});
can be made to both the methods. So, an ambiguity is there.
However, f(5) call can only be made for the first method. Since, 5 cannot be assigned to an array type.
So, the ambiguity only occurs when you are passing an array.
your compiler thinks that the method which takes variable arguments as an argument might be the same as an method which takes an array as an argument. i.e., it thinks there are duplicate methods with the same number of arguments, which contrays overloading rules.
public void m1(int[] arr){
}
public void m1(int...i){
}
are basically same.the only difference is var-args can accept any number of int variables
Method overloading is a feature of java that allows you to more then one method having the same name but different argument list
METHOD OVERLOADING IS ALSO KNOWN AS STATIC POLYMORPHISM
Argument list must be differ
1) Numbers of Parameters "not the same parameters"
2) Data type of parameters "does not have the same datatype of parameters"
3) Sequence of Datatype of parameters "its may be possible to change the sequence the of datatype."
The method overloading have the same argument then it must be the return type is different suppose
public class Test
{
public int printVal(int i,int j)
{
System.out.println("Integer Return type")
return i*j;
}
public double printVal(int i,int j)
{
System.out.println("Double Return type")
return i*j;
}
}
In the above class we have the two method one have int return type and another have double return type so that can be possible in method overloading.
Sequence of datatype it means
public class Test
{
public void display(String surname ,String name)
{
System.out.println("Surname = "+surname);
System.out.println("Name = "+name);
}
public void display(String name,String surname)
{
System.out.println("Name = "+name);
System.out.println("Surname = "+surname);
}
}
in above example we have two method as a same datatype but its sequence is differ this case is also called method overloading..
The method signature is the same. Thus overload is impossible. Change the method signature.
There's a definition of method signature in Java from wikipedia.
In the Java programming language, a method signature is the method name and the number and type of its parameters. Return types and thrown exceptions are not considered to be a part of the method signature.
In your code you have two methods with the same signature f(int[] value). Another function uses int... as an argument, but it's equivalent to int[].
Related
i made a small program for summing two numbers
if i used a void type method it will be like this
import java.util.Scanner;
public class NewClass {
public static void main(String[] args) {
Scanner input = new Scanner (System.in);
System.out.println("enter x");
int x = input.nextInt();
System.out.println("enter y");
int y = input.nextInt();
getSum(x,y);
}
public static void getSum(int x, int y)
{
int sum = x + y;
System.out.println("sum = " + sum);
} }
here if i used a method that returns a value i will get same output!
import java.util.Scanner;
public class NewClass {
public static void main(String[] args)
{
Scanner input = new Scanner (System.in);
System.out.println("enter x");
int x = input.nextInt();
System.out.println("enter y");
int y = input.nextInt();
int r = getSum(x,y);
System.out.println("sum = " + r);
}
public static int getSum(int x, int y)
{
int sum = x + y;
return sum;
}}
so i'm confused what is the benefit of returning a value if i can finish my program with a void type method?
The point is not to get the result but the way we get the result
consider
public static void getSum(int x, int y)
{
int sum = x + y;
System.out.println("sum = " + sum);
}
will print the output to the screen after calculation
but when we use return
as you have done it later
public static int getSum(int x, int y)
{
int sum = x + y;
return sum;
}
this function will respond back the sum. that sum can be stored in a variable
can be used afterwards
like in recursion
In small programs, you won't get the difference but while writing the big programs you have to make several functions which are being called several times and you may need the output of one function into other.
In that case, you will require return so that the output of one function can be used into other.
Hope this helps!!
I think the answer is that, if you're calling getSum() method with a return type in any other class.. you would get a returned value which can be used for further processing. .
Where as in void that's not possible... Hope this helps... Reply if any doubts..
I can understand why you have this question.
First of all, you should know that in real development, we do not use console logs.
System.out.Println();
This is used only for debugging and that too in very rare cases.
The whole point of a function is to take some input and convert to something else.
So,
public static int getSum(int x, int y) {
return x+y;
}
public static void main(String[] args) {
System.out.Println(getSum(5,10));
}
This is the better solution.
Best Regards,
Rakesh
When you use the keyword void it means that method doesn't return anything at all. If you declare a return type different to void at the method statement instead, that method must return obligatorily a valid value to the declared return type using the keyword return followed by a variable/value to send back to the class that called the method.
Defining methods here you have the java documentation for a method declaration
Answering your question, in small programs that work with primitive values it doesn't really matter but in complex program when you usually need to return specifics object types, i.e an ArrayList or actually an instance of a class you created you can't simply put it into a System.out.println and send it to the console, mostly you'll want to get something from a method and that something usually can be a more complex object than an integer or a string, the way to get that something is through the return type defined by the method's statement.
A common use of return types is when your method is static and it can't interact with the non-static instance variables of the class, this type of static methods usually get values from their arguments, do a certain kind of progress and then return a result that the method's caller can use.
Returning a value enables you to use that value in whichever way you want, including printing it or assigning it to variable for further processing. If on the other hand you print the value in the method and not return anything, i.e. making the method of type void, then that's all you can do with that method.
class TestOverride{
public int testVlaue(int a,int b){
return a + b;
}
public float testValue(int x,int y){
return x + y;
}
public float testValue(float x){
return x;
}
}
public class CodeTester{
public static void main(String a[]){
TestOverride objTest = new TestOverride();
System.out.println(objTest.testValue(2));
System.out.println(objTest.testValue(2,3));
}
}
Why is the output as follows?
2.0
5.0
It can also take the return type int instead of float for return value 5?
If you want to do overloading, you must change in method signature for one of these points:
Parameter types.
Number of parameters.
Order of the parameters declared in the method.
Return type is not considered one of the method signature so if you have two methods with same name and different return type you will get compilation error, it is not allowed to do that.
Last,in your code you have two methods with different name so there is no compilation error and you call the method which return float
see the different in vl and va
testVlaue
testValue
you call System.out.println(objTest.testValue(2,3)); the second one
In the following code the method array1 won't return the avarage because its return type is void.
I know what void means but can someone explain to me what is a void result type and how to make this method return the avarage:
public class JavaApplication4 {
public static void main(String[] args) {
int[] a = {1,2,3,4};
double result = array1 (a);
}
public static array1 (int[] b) {
double avarage;
int total = 0;
for (int x:b) {
total += x;
}
avarage = total / b.length;
return avarage;
}
The "result type" or "return type" is set in the function declaration. It just means what type of data is going to be returned after the function is called. Your function should look like:
public static double array1(int[] b) {
double average;
int total=0;
for(int x:b){
total +=x;
}
average = (double) total/b.length;
return average;
}
This will return the value of average after the function is done. So result will hold the same thing as the final value of average after the function completes.
You need to declare array1 as returning double. Change its declaration to:
public static double array1(int [] b ) {
A void function -- void array1(...) -- does not return a value.
Note that there's another error in your code:
avarage = total / b.length;
The above uses integer division, meaning that the result is truncated to integer, and only then converted to double. To fix, change the line to:
avarage = total / (double)b.length;
A return type of void means that a method doesn't return anything. This is useful when you want to preform an operation on an array, but there isn't any value associated with the operation. For example, say that you wanted to swap the first and last element in an array, you could write some method like this (ignore the necessary error checking)
public void swapArrayLocs(int[] swapping){
int temp = swapping[0];
swapping[0] = swapping[swapping.length - 1];
swapping[swapping.length - 1] = temp;
}
When you call this method, you're not expecting any sort of result from it, you're expecting your program to just take care of business and continue executing.
In this case, you actually want your array1(int[]) method to make its available to the rest of the program. You do this by specifying the return type of the function, which tells the rest of the program what type of information you expect that function to return. In your case, you'd do this by changing your method declaration to.
public static double array1(int[] b){
//the same method body
}
Note how in this case the word double is inserted after static. This tells the calling function that when it executes array1, the method will give back a value of type double. Contrast this with what you had before which said that the method would not give back any type of information.
public static void main(String args[])
{
double arr[] = {1,-6.3,9000,67.009,1.1,0.0,-456,6,23,-451.88};
ArrayList<Integer> List = new ArrayList<Integer>();
List.add(1);
List.add((int) -6.3);
List.add(9000);
List.add((int) 67.009);
List.add((int)1.1);
List.add((int)0.0);
List.add(-456);
List.add(6);
List.add(23);
List.add((int)451.88);
}
public static int ArrayListMax(ArrayList List)
{
for (int i=0; i<List.size(); ++i)
{
System.out.println(List.get(i));
}
The error is in:
public static int ArrayListMax(ArrayList List)
This is probably a very nooby mistake, but I'm new to Java so forgive me.
Any help please?
Thank you.
EDIT:
I want the ArrayListMax method to print the size of the List!
Assuming you are trying to get the maximum value in your List in the method arrayListMax, you need to return an integer in accordance with your method signature, which the error is telling you
This method must return a result of type int
Instead of printing all the values in the list, you could do:
public static int arrayListMax(List<Integer> List) {
return Collections.max(list);
}
Use Java Naming Conventions. Method & variable names begin with a lowercase letter. Using this approach helps avoid confusion between instances & types (e.g. in the case of List in the main method).
Either add a return statement to your ArrayListMax() method that returns an int or change the method signature from public static int to public static void. And add a closing } to the method too.
Also, you shouldn't use List as a name for the argument to that method because it's the name of an interface that you're actually importing in this code. The convention in java is for variable names and method names to begin with a lowercase letter (camel case) and class names to begin with an uppercase letter (pascal case).
Problem is in this method:
public static int ArrayListMax(ArrayList List)
{
for (int i=0; i<List.size(); ++i)
{
System.out.println(List.get(i));
}
}
You are not returning from this method, you must return of type int to solve that error. Since you have specified int as return type.
You created the function as returning a result
public static int ArrayListMax(ArrayList List)
In order to remove your problem, if you need not return anything, write
public static void ArrayListMax(ArrayList List)
void means you do not want the method to return a result.
if you are just printing out values the signature should be public static void ArrayListMax(ArrayList List).
First, you are not returning anything with ArrayListMax method even if you have set its retun type to int. Either return void as you are directly printing from the list.
Second, even if you set the correct return type to ArrayListMax method, you will still need to call that method in the main method as below:
public static void main(String args[])
{
double arr[] = {1,-6.3,9000,67.009,1.1,0.0,-456,6,23,-451.88};
ArrayList<Integer> List = new ArrayList<Integer>();
List.add(1);
List.add((int) -6.3);
List.add(9000);
List.add((int) 67.009);
List.add((int)1.1);
List.add((int)0.0);
List.add(-456);
List.add(6);
List.add(23);
List.add((int)451.88);
YourClassNameInWhichMethodIs.ArrayListMax(List);
}
You should use a return statement as Methode is returning an int type value.
This question already has answers here:
What is the ellipsis (...) for in this method signature?
(5 answers)
Closed 7 years ago.
I was looking through some code and saw the following notation. I'm somewhat unsure what the three dots mean and what you call them.
void doAction(Object...o);
Thanks.
It means that this method can receive more than one Object as a parameter. To better understating check the following example from here:
The ellipsis (...) identifies a variable number of arguments, and is
demonstrated in the following summation method.
static int sum (int ... numbers)
{
int total = 0;
for (int i = 0; i < numbers.length; i++)
total += numbers [i];
return total;
}
Call the summation method with as many comma-delimited integer
arguments as you desire -- within the JVM's limits. Some examples: sum
(10, 20) and sum (18, 20, 305, 4).
This is very useful since it permits your method to became more abstract. Check also this nice example from SO, were the user takes advantage of the ... notation to make a method to concatenate string arrays in Java.
Another example from Variable argument method in Java 5
public static void test(int some, String... args) {
System.out.print("\n" + some);
for(String arg: args) {
System.out.print(", " + arg);
}
}
As mention in the comment section:
Also note that if the function passes other parameters of different
types than varargs parameter, the vararg parameter should be the last
parameter in the function declaration public void test (Typev ... v ,
Type1 a, Type2 b) or public void test(Type1 a, Typev ... v
recipientJids, Type2 b) - is illegal. ONLY public void test(Type1 a,
Type2 b, Typev ... v)
It's called VarArgs http://www.javadb.com/using-varargs-in-java. In this case, it means you can put multiple instances of Object as a parameter to doAction() as many as you wants :
doAction(new Object(), new Object(), new Object());