Java Out Of Bounds Error
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Index Out Of Bounds Exception C#
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How To Solve Array Index Out Of Bounds Exception In Java
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Arrayindexoutofboundsexception Java
The error that I am getting Exception in thread "main" java.lang.ArrayIndexOutOfBoundsException: 610 at Fib.sorted(Fib.java:67) at Fib.main(Fib.java:17) My code public class Fib { public static void main(String args[]) { System.out.println(Arrays.toString( fiblist) ); System.out.println(Fib.add()); System.out.println(Fib.square()); System.out.println(Fib.reversal()); System.out.println(Fib.sorted()); } public static int fiblist[] = {1,1,2,3,5,8,13,21,34,55,89,144,233,377,610,987,1597,2584,4181,6765}; public static int fiblen = fiblist.length; public Fib() { // Do nothing } public static ArrayList
here for a quick overview of the site Help Center Detailed answers to any questions you might have Meta Discuss the workings and policies of this site About Us array index out of bounds c++ Learn more about Stack Overflow the company Business Learn more about hiring developers array index out of bounds exception 0 or posting ads with us Stack Overflow Questions Jobs Documentation Tags Users Badges Ask Question x Dismiss Join the Stack Overflow array index out of bounds exception processing Community Stack Overflow is a community of 4.7 million programmers, just like you, helping each other. Join them; it only takes a minute: Sign up Out of bounds error on a simple String http://stackoverflow.com/questions/26149182/how-to-fix-array-index-out-of-bounds-error array [SOLVED] up vote -2 down vote favorite I am trying to make a method that will divide a String up into groups of 4, and I keep getting an out of bounds error. I can't understand where it is coming from, is there an easier way to do this, or what is causing the error? EDIT: SOLVED bad brackets means int MAX isn't assinged the correct http://stackoverflow.com/questions/26209631/out-of-bounds-error-on-a-simple-string-array-solved value The relevant code public void ColumnCode(String plaintext) { System.out.println("Please enter the number of columns you want"); int columns = slave.getNum(); int MAX = (plaintext.length()-1/columns)+1; //number of times to iterate the for loop String[] groups = new String[MAX]; for(int i = 0; i < MAX; i++) { if((4*i)+4 > plaintext.length()) groups[i] = plaintext.substring(4*i); else groups[i] = plaintext.substring(4*i, 4*(i+1)); System.out.println(groups[i]); } } Example error message: Now please enter the plaintext to be encrypted: THIS IS A MESSAGE Please enter the number of columns you want 4 THIS IS A ME SSAG E Exception in thread "main" java.lang.StringIndexOutOfBoundsException: String index out of range: -3 at java.lang.String.substring(Unknown Source) at Cryptopher.ColumnCodeClass.ColumnCode(ColumnCodeClass.java:26) at Cryptopher.mainFile.encrypt(mainFile.java:86) at Cryptopher.mainFile.runTheProgram(mainFile.java:55) at Cryptopher.mainFile.main(mainFile.java:36) java runtime-error share|improve this question edited Oct 6 '14 at 3:35 asked Oct 6 '14 at 3:24 spyr03 275214 You are getting the index of of bounds error because on the last iteration you are suppose to have 4 elements in your array, but your array is only populated with 1 element ("E"). –orgtigger Oct 6 '14 at 3:36 add a comment| 2 Answers 2 active oldest votes up vote 1 down vote accepted The error is caused by this line: int MAX = (p
I getting array index out of bounds exception?say I have a method which among its arguments takes an int. I use that int in the method to index into an array which is an instance variable.in the first call https://www.quora.com/Why-am-I-getting-array-index-out-of-bounds-exception of that method, intarg is zero. Why am I getting an exception when the array is guaranteed to have a 0th element? private void someMethod(somearg, somearg, intarg) {....................................................................................myArray[intarg] += someint;}UpdateCancelAnswer Wiki10 Answers Chandra Mohan, Full Time Java professional now!Written 170w agoAs Apurva Nandan mentioned in the comment, looks to be a problem in the initialization code or hang on.. it can't be initialization error since then we would have had NullPointer.Do out of make sure the intarg is received as 0 and not any other number. Also make sure the intarg is not modified before reaching that part of the code in the method. (Parameters are not readonly and can be modified.)12.3k Views · View UpvotesRelated QuestionsMore Answers BelowHow do I get rid of the runtime exception saying ArrayIndexOutOfBound at the line on which I declared a jagged array?What does "exception in thread main java.lang.Arrayindex out of bounds out of bounds" means in java?What does this mean when I get an exception and I get null?Why am I getting "cannot find symbol"?Why am I getting Stub exception in Appium? Kaustubh SahaWritten 114w agoCheck the following :1. Is the int parameter arg being reinitialized somewhere in the method ? if not, may be consider making it final so as to avoid a red herring2. An array is not guaranteed to have a 0th element. Its perfectly legal to declare an array of size 0. In that case even arr[0] will give IndexOutofBoundsExceptionFor example, the following code will throw java.lang.ArrayIndexOutOfBoundsException :int[] arr = new int[0]; System.out.println(arr[0]);3. Consider enabling assertions and adding an assertion checking if the length of myArray is >= arg14.3k Views · View Upvotes Amar Chaitanya, Java developerWritten 49w agoArray index out of bounds exception occurs, for example, when you have 3 elements in the array and you are trying access element 4th element. This usually occurs when we forget that the indexing starts from 0. Remeber that you can start the loop and end at any number when you want execute a specific set of lines for certain number of times. But when you are looping to access elements with the indexing variable you should be careful about the