Not sure where I'm going wrong with this. I've asked someone in my class and they said there should be an argument with "toonRijSterren". when I do this I just get more errors, could someone have a look and tell me where I'm going wrong?
public static void main(String[] args) {
int aantal = 0;
toonRijSterren(aantal);
toonSterrenVierkant(aantal);
}
public static void toonRijSterren(int mpAantal) {
while (mpAantal < 6) {
System.out.print(" * ");
mpAantal++;
}
}
public static void toonSterrenVierkant(int mpAantal) {
for (int mpAatal = 0; mpAantal < 6; mpAantal++) {
System.out.println(toonRijSterren());
}
}
ther error line is in the brackets of the last toonRijSterren());
toonRijSterren is void method which means it does not return any value and therefore you can not put it inside System.out.println() or you can not assign it to some variable.
toonRijSterren expects an int argument which you have missed while calling it.
Given below is an example of how you should call toonRijSterren:
public static void toonSterrenVierkant(int mpAantal) {
for (int mpAatal = 0; mpAatal < 6; mpAatal++) {
toonRijSterren(mpAantal);
}
}
You are not passing the argument when you call your method.
Try this:
System.out.println(toonRijSterren(mpAatal));
First of all, your function toonRijSterren takes an int type parameter (according to its declaration), so you need to pass to it another argument. For example:
toonRijSterren(mpAantal)
Second, the function toonRijSterren returns void. That means, it just does an operation (in this case, printing) without returning anything. What you're trying to do is to use its return value (which doesn't exist) as an argument to System.out.println, which causes an error (because println expects an argument of some type).
You could achieve what I think you're trying to do with the line:
toonRijSterren(mpAantal);.
The function itself prints the values, so the println here is unnecessary and causes an error.
You are missing the parameter in your toonSterrenVierkant() function where you calling toonRijSterren.
Here is the corrected version of your code:
public static void toonSterrenVierkant(int mpAantal) {
for (; mpAantal < 6; mpAantal++) {
toonRijSterren(mpAatal);
}
}
As your methed toonSterrenVierkant(int mpAantal) has a int parameter, you must pass an int value as an argument in the last toonRijSterren(). For example, replace the line System.out.println(toonRijSterren()); with System.out.println(toonRijSterren(1));
Related
I do not understand the details of using an array as a parameter:
I have created an int array and a method and I do not understand why it's possible to rename the parameter as seen below from "note" to "veraenderung".
How does Java deal with parameters? Do I need a parameter even if I call the method(note)?
public class ArrayParameter {
public static void main(String[] args) {
int[] note = {3,2,1,4,5}; //Array
int[] note2 = {3,2,1,4,5};
korrektur(note);
korrektur2(note2);
}
//Der Methode "korrektur" wird der int Array (note) übergeben!
public static void korrektur (int note[]) {
for (int i = 0; i<note.length; i++) {
//Sobald die for Schleife aufgerufen wird, werden die jeweiligen Werte um -1 reduziert
note[i]-=1;
System.out.println(note[i]);
}
}
//int "veraenderung" ist ein Parameter
public static void korrektur2 (int veraenderung[]) {
for (int i = 0; i<veraenderung.length; i++) {
//Sobald die for Schleife aufgerufen wird, werden die jeweiligen Werte um -1 reduziert
veraenderung[i]-=1;
System.out.println(veraenderung[i]);
}
}
}
i do not understand why it is possible to rename the parameter
korrektur and korrektur2 are just 2 different methods each of them having 2 independent signatures. They are not related whatsoever (even if the programmer can name them alike, just as you did);
The name of the parameters are just local names useful in the method block;
How do java deal with parameter? Do i need a parameter even if i call the method(note)?
Those methods parameters are mandatory ones. Hence you need to insert them in your method call.
I will try to break it down as small as possible for you to understand :)...
When you creating the method. example: public static void korrektur2 (int veraenderung[]) the int veraenderung[] can be named whatever you want...the main thing is that you put the "int []" somewhere within the brackets so that you tell java that "hey my method take an integer array".
The name veraenderung itself is only used locally (within that method) as a reference to tell java whatever you parse into that method will act.
For example when you call the method above and said korrektur(note); . note was declared as an array before, so when you put 'note' inside the method korrektur it is the same as replacing veraenderung with note. In other words java replaces your local variable with the one you passed into the method.
And to answer your question....for now...since you created the method which takes an integer array...YES you must always "put an integer array inside" it when calling the method. I.e. every time you call the method korrektur2 or korrekturand you do not put an integer array inside it, Java would be like "hey I was created to take an integer array and do stuff with it...why are you leaving me empty ?!?!?"
I hope this solved your questions :)
If you have seen the .class file, you'll find the parameters are the same.
you should understand the conception 'vars Scope' (may be this). Parameter 'paramArrayOfInt' have the influence korrektur(), then it be destroied. In next method ,it just new one. It's my understanding. May be wrong, welcome to discuss.
public class ArrayParameter
{
public static void main(String[] paramArrayOfString)
{
int[] arrayOfInt1 = { 3, 2, 1, 4, 5 };
int[] arrayOfInt2 = { 3, 2, 1, 4, 5 };
korrektur(arrayOfInt1);
korrektur2(arrayOfInt2);
}
public static void korrektur(int[] paramArrayOfInt)
{
for (int i = 0; i < paramArrayOfInt.length; i++)
{
paramArrayOfInt[i] -= 1;
System.out.println(paramArrayOfInt[i]);
}
}
public static void korrektur2(int[] paramArrayOfInt)
{
for (int i = 0; i < paramArrayOfInt.length; i++)
{
paramArrayOfInt[i] -= 1;
System.out.println(paramArrayOfInt[i]);
}
}
}
I'm just new to Java OOP, and I hardly understand about the class and stuff. I tried to write some code to understand but I didn't succeed. Here is my code, I expected it to return the number of eggs but I don't know why it returns nothing.
class EggCounter {
private int egg;
{
egg = 0;
}
public void eggAdd()
{
egg = egg + 1;
}
public void eggBreak()
{
egg = egg - 1;
}
public void eggAddDozen()
{
egg = egg + 12;
}
public int getEgg()
{
return egg;
}
}
public class EggTest
{
public static void main(String[]args)
{
EggCounter egg = new EggCounter();
egg.eggAdd();
egg.eggAddDozen();
egg.eggBreak();
egg.getEgg();
}
}
It does return 12. Replace the egg.getEgg(); line in your main method with System.out.println(egg.getEgg()); and you will notice it prints 12.
It is returning, it's just that you do nothing with the return value of getEgg. What you need to do is store it into the variable or do something with it. return <value> only returns the given value to the callee, you must store it to use it. Example:
int eggCount = egg.getEgg();
System.out.println(eggCount);
Here, the assignment of eggCount calls egg.getEgg(). The call resolves when the number of eggs is returned, which assigns the return value to eggCount. Finally, it will print out eggCount. If you need the result of egg.getEgg() later, you can simply just output the following:
System.out.println(egg.getEgg());
How this works is the method egg.getEgg() is called. The return value is then resolved, which is passed into the print statement. This gets rid of storing it into a variable you can use later.
when I was writing my program I came across a problem with arrays and can't find any explanations on how to fix it.
The problem i ran into is how i should go about using an array from the main a method and passing it to a method, but only getting one value as a return rather than the whole array again.
note: the way i have my program set up the array is filled with random numbers every time it is ran.
im not too sure if you can do this or not, but
The way i am currently calling a method is by doing as follows:
"declared variable (not array)" = "name of method" ("the actual array name")
totalSum = sumTotal (randomArray) // example
The method I'm trying to call:
public static int[] sumTotal (int[] totals) {
int total = 0;
for (int i = 0; i < totals.lenght; i++) {
total += totals[i];
}
return total;
}
it keeps giving me a "cannot find symbol
symbol : method sumTotals(int[])
location: class oneDimensionArraysNew"
error when i try to compile,
im not too sure how i would go about fixing this, I appreciate any and all help!!
Here you say, the return value is array of int : public static int[] sumTotal...
Just change it to : public static int sumTotal...
public static int[] sumTotal(int[] totals {
...
}
The return type should be int, not int[].
public static int sumTotal(int[] totals {
...
}
Need help on calling a method from main class.
I need to call a method, thus I made an object to handle it.
below I quote my main method
public static void main(String[] args) {
// TODO code application logic here
SLRatio sl= new SLRatio();
sl.clustering(apa);
}
and here's the method I need to call
public class SLRatio {
public static String [][]clustering(String[][]apa) {
System.out.println("Cluster 1");
int a = apa.length/3;
String [][] cluster1=new String [a][apa[0].length];
for (int i =0; i<a; i++) {
for (int j=0;j<apa[0].length;j++) {
cluster1 [i][j] = apa[i][j];
}
}
for (int b = 0; b < cluster1.length; b++) {
for (int c = 0; c < cluster1[0].length; c++) {
System.out.print(cluster1[b][c] + "\t");
}
System.out.println("");
}
System.out.println("\n");
return cluster1;
}
}
and I got error message:
"Cannot find symbol,Accessing static method clustering"
What can I do to solve it? I have tried to change the syntax but it didn't work.
Thank you so much.
you didn't define method Allocation() in SLRatio
Note: static method should be called with classname (to avoid confision between instance method and static)
If it is static method, you don't need to call it through instance.
SLRatio .clustering(...);
should be enough.
And it seems you forgot to implement Allocation method.
Another suggestion, java naming convention, method name starts with small case letters.
Do not use static unless you are sure it is appropriate.
This is a popular programming error, partially because eclipse keeps on suggesting to make variables and methods static when they cannot be accessed. But usually, this is not the correct solution. While it fixes the compilation problem, it often breaks the application logic.
Right now, your problem probably is that apa has type String[][], but you are passing a String[] parameter to it. So it cannot be compiled, because there is no method clustering(String[] args).
Seriously, you need to learn more Java basics. Maybe from a book.
Something is very wrong here...
public class evenness {
public static Boolean isEven (Integer i) {
return (i % 2) == 0;
}
public static void main(String[] args) {
if (isEven(Integer i)) { //something wrong on this line.
System.out.print("YAY!");
}
}
}
Please help me sort it out!
You are supposed to give an Integer argument to the function isEven, for example 3 or 125.
I'm pretty sure that your IDE is telling you that it can't find the variable Integer. So you need to give a variable. The argument type is only required in the method definition.
For example:
int number = 4;
if(isEven(number)){
...
}
or more directly
if(isEven(4)){
...
}
Change
if (isEven(Integer i)) {
to something like
int i = 5;
if (isEven(i)) {
or
if (isEven(5)) {
You need to pass an integer to the method.
Just provide an argument where you call your method in main();
public class evenness {
public static Boolean isEven (Integer i) {
return (i % 2) == 0;
}
public static void main(String[] args) {
if (isEven(36)) { //something wrong on this line.
System.out.print("YAY!");
}
}
}
Following Java coding convention by changing the class name to Evenness.
public class Evenness
The method isEven(...) should accept int and return boolean. Both are primitive data types. This will make the program runs faster than using their wrappers. A wrapper is a reference type (object) that wraps a primitive type. For example, Integer wraps int and Boolean wraps boolean.
public static boolean isEven (int i) {
Send an argument to isEven(...)
if (isEven(2)) { // this line is now OK.