Dereferenced Java Error
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Double Cannot Be Dereferenced Java
this "Int cannot be dereferenced" error when trying to compile and I'm not sure what the problem is. The error is specifically happening in my if statement at the bottom, where it says "equals" is an error and "int cannot be dereferenced." Hope to get some assistance as I have no idea what to do. Thank you in advance! public class Catalog { private Item[] list; private int size; // Construct an empty catalog with the int cannot be dereferenced tostring specified capacity. public Catalog(int max) { list = new Item[max]; size = 0; } // Insert a new item into the catalog. // Throw a CatalogFull exception if the catalog is full. public void insert(Item obj) throws CatalogFull { if (list.length == size) { throw new CatalogFull(); } list[size] = obj; ++size; } // Search the catalog for the item whose item number // is the parameter id. Return the matching object // if the search succeeds. Throw an ItemNotFound // exception if the search fails. public Item find(int id) throws ItemNotFound { for (int pos = 0; pos < size; ++pos){ if (id.equals(list[pos].getItemNumber())){ //Getting error on "equals" return list[pos]; } else { throw new ItemNotFound(); } } } } java int bluej share|improve this question edited Jul 19 at 0:13 Sotirios Delimanolis 154k25246361 asked Oct 1 '13 at 6:08 BBladem83 183129 2 You're trying to using a int where an Integer, Number or Object is expected...int does not have any methods –MadProgrammer Oct 1 '13 at 6:09 add a comment| 5 Answers 5 active oldest votes up vote 9 down vote accepted id is of primitive type int and not an Object. You cannot call methods on a primitive as you are doing here : id.equals Try replacing this: if (id.equals(list[pos].getItemNumber())){ //Getting error on "equals" with if (id == list[pos].getItemNumber()){ //Getting error on "equa
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 numerous times, but int cannot be dereferenced meaning I am still confused about why I received the error. I am very
Int Cannot Be Dereferenced Compareto
new to Java, this is my first and last programming class. This is the first time I have seen this
Int Cannot Be Dereferenced Length
error 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 http://stackoverflow.com/questions/19109131/int-cannot-be-dereferenced-in-java thru the array and find 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 < https://coderanch.com/t/536655/java/java/int-dereferenced-error 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 (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? B
This Site Careers Other all forums Forum: Java in General What is meant by dereferencing? prabal saha Greenhorn Posts: 4 posted 15 years ago hi there, while compiling a program, i https://coderanch.com/t/367369/java/java/meant-dereferencing got this error: " byte cannot be dereferenced" please let me know what actually dereferencing is? prabal. Madhav Lakkapragada Ranch Hand Posts: 5040 posted 15 years ago Dereferensing means setting the reference to NULL. This way you are no longer referencing any obj. This is one way. Could you post your code..... regds. - satya Take a Minute, Donate an Hour, Change a cannot be Life http://www.ashanet.org/workanhour/2006/?r=Javaranch_ML&a=81 prabal saha Greenhorn Posts: 4 posted 15 years ago thanks Madhav, i got the point that was causing this eror.As for the code my approach was wrong. Frank Carver Sheriff Posts: 6920 posted 15 years ago 1 I don't agree with the explanation here. "dereferencing" actually refers to the process of getting or setting the value referred to by cannot be dereferenced a reference. For example, If I say String s = "hello" then s is a reference to the actual characters. In this case I can dereference s to get the characters, but if I say, for example String s = null; then I can't dereference s, because it does not refer to any characters - it is null! In the case of the initial question, I would imagine that the code was trying to call a method on a byte, and as byte is a primitive type which contains its own value, and not a reference to an Object, it could not be dereferenced. For example: byte b = 45; System.out.println(b.toString()); will not work (if you need to call a method, you should have used the object wrapper class Byte, instead), and neither will byte b; b[7] = 23; where the [] have inadvertently omitted from an array declaration. Read about me at frankcarver.me ~ Raspberry Alpha Omega ~ Frank's Punchbarrel Blog rehan hamid Greenhorn Posts: 16 posted 15 years ago In general, Reference is an address to some object/variable, While getting or setting value for that variabl