Int Cannot Be Dereferenced Error Java
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Certification Databases Caching Books Engineering Languages Frameworks Products This Site Careers Other all forums Forum: Beginning Java int cannot be dereferenced error Collete Williams Greenhorn Posts: 4 posted 5 years ago I know this error has been written int cannot be dereferenced tostring numerous times, but I am still confused about why I received the error. I
Int Cannot Be Dereferenced Meaning
am very new to Java, this is my first and last programming class. This is the first time I have seen this error int cannot be dereferenced compareto and lost. I know it has something to do with the int and string. I am confused all together how it works. Code is suppose to accept an integer from user, go thru the array and find int cannot be dereferenced length the element of that array and display in a textfield. It also has to handle 2 specific exceptions. Can someone please tell me what it is suppose to look like. Thank you.. int cannot be dereferenced-line 53 //import java.lang.*; public class showElement implements ActionListener{ public void actionPerformed(ActionEvent e){ randomNumber = new int [100]; for (int x = 0; x < randomNumber.length; x++) randomNumber[x] = (int)(Math.random() * 1000); String inputString; inputString=valueField.getText(); try{ valueField.setText(randomNumber[Integer.parseInt(inputString)].toString());//error section } catch
Int Cannot Be Dereferenced Equals Java
(IndexOutOfBoundsException ex){ valueField.setText("Out of bounds"); } catch (NumberFormatException ex){ valueField.setText("Not a Integer."); } } } Kurt Van Etten Ranch Hand Posts: 98 posted 5 years ago 1 Hi Collete, and welcome to the Ranch! The error you're getting, in line 17 of the code excerpt you posted, is because you're trying to call the toString() method on an int (since that's what's stored in the array), and int is a primitive type. There are several ways you could convert the int to a String: for example, you could concat it with an empty string to have it implicitly converted, or you could use the static Integer.toString() method to explicitly convert it . Collete Williams Greenhorn Posts: 4 posted 5 years ago I read up on on what it means to "concat it with an empty string" and "static Integer.toString() method" but I'm sorry I am still confused. Originally I had major help with that line but it looks like it did not work. Are you saying this line: valueField.setText(randomNumber[Integer.parseInt(inputString)].toString()); needs to changed to one of the options? Been trying for 2 weeks to get this working and I am burned out majorly and have gone brain dead. Is it possible for you to show me if not maybe a example. Campbell Ritchie Sheriff Posts: 50555 82 posted 5 y
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Int Compareto Java
like you, helping each other. Join them; it only takes a minute: Sign up Error of int cannot be dereferenced? up vote 0 down vote favorite I am getting an error with this constructor, and i have https://coderanch.com/t/536655/java/java/int-dereferenced-error no idea how to fix? I am a beginner at java. This is from an example exercise that i was trying to learn: /** * Create an array of size n and store a copy of the contents of the * input argument * @param intArray array of elements to copy */ public IntArray11(int[] intArray) { int i = 0; String [] Array = new String[intArray.length]; for(i=0; i
need some help in my programme. I have an error message: "error: int cannot be dereferenced" at line 134. Could you help me solve https://www.daniweb.com/programming/software-development/threads/379119/error-error-int-cannot-be-dereferenced this problem, as I do not know what it means? Many https://bytes.com/topic/java/answers/771208-int-cannot-dereferenced-error-message thanks. /** * @(#)PrizeCollection.java * * PrizeCollection application * * @author * @version 1.00 2011/8/25 */ import javax.swing.JOptionPane; import java.lang.*; import java.util.*; public class PrizeCollection { public static void main (String[]args) { // Declare variables and arrays which will be used in the programme cannot be String[] Description = new String [3]; String[] Color = new String [3]; int[] Value = new int [3]; //Call the Menu optionMenu(Description,Color,Value); } //Method for the menu that will appear each time a task is completed public static void optionMenu(String Description[], String Color[], int Value[]) { int option = 0; while (option!=1 || option!=2 || option!=3 int cannot be || option!=4){ option = Integer.parseInt(JOptionPane.showInputDialog("Please choose an option: \n" + "1 - Enter the details of a prize \n" + "2 - Print the details stored for all prizes \n" + "3 - Search for a prize with a particular value or by description \n" + "4 - Quit")); if (option==1) enterDetails(Description,Color,Value); if (option==2) printDetails(Description,Color,Value); if (option==3) searchDetails(Description,Color,Value); if (option==4) System.exit(0);} System.exit(0); } //Method to enter details for each prize public static void enterDetails(String Description[], String Color[], int Value[]){ for (int i=0; i<3; i++){ Description [i] = JOptionPane.showInputDialog (null, "Please enter the description of the prize: "); Color [i] = JOptionPane.showInputDialog (null, "Please enter the color of the prize: "); Value[i] = Integer.parseInt(JOptionPane.showInputDialog(null, "Please enter the value of the prize: ")); } optionMenu(Description,Color,Value); } // Method to search prizes public static void searchDetails(String Description[], String Color[], int Value[]) { // A Menu if needed to choose between search by description or by value int choice = 0; while (choice!=1 || choice!=2 || choice!=3){ choice = Integer.parseInt(JOptionPane.showInputDia
your question and get tips & solutions from a community of 418,570 IT Pros & Developers. It's quick & easy. int cannot be dereferenced error message P: 6 JPane Expand|Select|Wrap|Line Numbers packagebankobject; importjava.io.*; importjavax.swing.JOptionPane; classNumber{ StringstudentName; StringteacherName; int[]Number; publicNumber(){ studentName=JOptionPane.showInputDialog("PleaseEnterstudentname"); teacherName=JOptionPane.showInputDialog("PleaseEntertheteachername"); intnumNumber=Integer.parseInt(JOptionPane.showInputDialog("Pleaseenterthenumberofnumbers")); Number[]obj=newNumber[numNumber]; for(i=0;i