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What is a StringIndexOutOfBoundsException? How can I fix it?
(1 answer)
Closed 3 years ago.
I am setting up a method that turn a infix string into a postfix equation with a custom LinkStack.
I have tried to to check if the charAt(i) was null and a if statement to check if i is greater than exp.length() but neither worked.
public static String infixToPostfix(String exp)
{
// make variable
String result = new String("");
int temp = 0;
LinkedStack stack = new LinkedStack();
for (int i = 0; i<exp.length(); ++i)
{
char c = exp.charAt(i);
if(Character.isDigit(c))
{
int n = 0;
//extract the characters and store it in num
while(Character.isDigit(c))
{
n = n*10 + (int)(c-'0');
i++;
c = exp.charAt(i); //exception occurs
System.out.println(n);
}
i--;
//push the number in stack
stack.push(n);
//System.out.println(stack.size() + ", Stack size");
}
// If ( push it to the stack.
if (c == '(')
stack.push(c);
// If ) pop and output from the stack
// until an '(' is encountered.
else if (c == ')')
{
while (!stack.isEmpty() && stack.peek() != '(')
result += stack.pop();
if (!stack.isEmpty() && stack.peek() != '(')
return "Invalid Expression"; // invalid expression
else
stack.pop();
}
else // an operator is encountered
{
while (!stack.isEmpty() && pre(c) <= pre((char) stack.peek()))
result += stack.pop();
stack.push(c);
}
}
// pop all the operators from the stack
while (!stack.isEmpty())
result += stack.pop();
String temp2 = stack.print();
System.out.println(temp2);
return result;
}
I expect the output to be 469 645 + if the input is 496+645 but the actual output is java.lang.StringIndexOutOfBoundsException: String index out of range: 7.
while(Character.isDigit(c))
{
n = n*10 + (int)(c-'0');
i++;
c = exp.charAt(i); //exception occurs
System.out.println(n);
}
You aren't length checking here, so you readily parse right off the end of the string.
while(i < exp.length() && Character.isDigit(c))
{
n = n*10 + (int)(c-'0');
if (++i < exp.length()) {
c = exp.charAt(i); //exception occurs
}
System.out.println(n);
}
Note: I'd cache the length because of how many times you use it, but that's not the cause of your problem.
Note, however, that this is cleaner code style:
public class Foo {
public static void main(String[] args) {
String myString = "12345";
int index = 0;
for (char c: myString.toCharArray()) {
System.out.printf("Char at %d == %c\n", index, c);
++index;
}
}
}
Notice the for-loop. I didn't do your calculations or break out or anything, but this is a cleaner way.
You can also do...
for (int index = 0; index < exp.length(); ++index) {
char c = exp.charAt(index);
if (!Character.isDigit(c)) {
break;
}
// Do other stuff here.
}
There are a variety of other ways to structure your code. Your while loop is awkward.
The rest of the code is working perfectly but I cannot figure out how to prevent punctuation from being translated.
public class PigLatintranslator
{
public static String translateWord (String word)
{
String lowerCaseWord = word.toLowerCase ();
int pos = -1;
char ch;
for (int i = 0 ; i < lowerCaseWord.length () ; i++)
{
ch = lowerCaseWord.charAt (i);
if (isVowel (ch))
{
pos = i;
break;
}
}
if (pos == 0 && lowerCaseWord.length () != 1) //translates if word starts with vowel
{
return lowerCaseWord + "way"; // Adding "way" to the end of string
}
else if (lowerCaseWord.length () == 1) //Ignores words that are only 1 character
{
return lowerCaseWord;
}
else if (lowerCaseWord.charAt(0) == 'q' && lowerCaseWord.charAt(1) == 'u')//words that start with qu
{
String a = lowerCaseWord.substring (2);
return a + "qu" + "ay";
}
else
{
String a = lowerCaseWord.substring (1);
String b = lowerCaseWord.substring (0,1);
return a + b + "ay"; // Adding "ay" at the end of the extracted words after joining them.
}
}
public static boolean isVowel (char ch) checks for vowel
{
if (ch == 'a' || ch == 'e' || ch == 'i' || ch == 'o' || ch == 'u' || ch == 'y')
{
return true;
}
return false;
}
}
I need the translation to ignore punctuation. For example "Question?" should be translated to "estionquay?" (question mark still in the same position and not translated)
As Andreas said, if the function is expecting only one word, it should be the responsibility of the calling function to ensure there's no full sentence or punctuation being passed to it. With that said, if you require the translator to handle this, you need to find the index of the string where the punctuation or non-letter character occurs. I added in a main method to test the function:
public static void main(String[] args) {
System.out.println(translateWord("QUESTION?"));
}
I added a loop into the qu case to find the punctuation being input, the two checks are to see if the character at position i is inside the range of a - z. The sub-string then only goes to the point where the punctuation is found.
int i;
for (i = 0; i < lowerCaseWord.length(); i++) {
if(lowerCaseWord.charAt(i) > 'z' || lowerCaseWord.charAt(i) < 'a') {
break;
}
}
String a = lowerCaseWord.substring (2, i);
String b = lowerCaseWord.substring(i);
return a + "qu" + "ay" + b;
This may need some tweaking if you're worried about words with hyphens and whatnot but this should put across the basic idea.
Here's the output I received:
$javac PigLatintranslator.java
$java -Xmx128M -Xms16M PigLatintranslator
estionquay?
My program works fine, but I'm getting some irregular spaces in the output. For example, if the input is 44 * 5 + 6 the output is 44<2 spaces>5<1 space>*<1space>6<no space>+. I tried fiddling with all the lines of code that are adding to the String postfix, but to no avail. I'd like the output to be of the form: operand<1 space>operand<1 space>operator (i.e. 1 space between operands and operators."
Here's my code:
import java.util.*;
public class PostfixConversion {
public static void main(String args[]) {
System.out.print("Enter an expression: ");
String infix = new Scanner(System.in).nextLine();
String postfix = convertToPostfix(infix);
System.out.println(postfix);
//System.out.println("The result of calculation is: " + postfixEvaluate("23+"));
}
//converts infix expression into postfix expression
public static String convertToPostfix(String infixExp) {
String postFix = "The Postfix Expression is: ";
Stack<Character> stack = new Stack<Character>();
char character = ' ';
for(int i = 0; i < infixExp.length(); i++)
{
character = infixExp.charAt(i);
//determine if character is an operator
if(character == '*' || character == '-' || character == '/' || character == '+')
{
//postFix += " ";
while(!stack.empty() && precedence(stack.peek(), character))
postFix += stack.pop();
stack.push(character);
} else if(character == '(') {
stack.push(character);
} else if(character == ')') {
while(!stack.peek().equals('(') && !stack.isEmpty())
postFix += stack.pop();
if(!stack.isEmpty() && stack.peek().equals('('))
stack.pop(); // pop/remove left parenthesis
} else
postFix += character;
}
while(!stack.empty()) //add the remaining elements of stack to postfix expression
{
if(stack.peek().equals('('))
{
postFix = "There is no matching right parenthesis.";
return postFix;
}
postFix += stack.pop();
}
return postFix;
}
public static boolean precedence(char first, char second) {
int v1 = 0, v2 = 0;
//find value for first operator
if(first == '-' || first == '+')
v1 = 1;
else if(first == '*' || first == '/')
v1 = 2;
//find value for second operator
if(second == '-' || second == '+')
v2 = 1;
else if(second == '*' || second == '/')
v2 = 2;
if(v1 < v2)
return false;
return true;
}
First remove all whitespaces from input, so that they don't destroy your formatting:infixExp = infixExp.replaceAll("\\s",""); and then add whitespaces where needed.
I've been having trouble with this assignment:
Given a string, replace the first occurrence of 'a' with "x", the second occurrence of 'a' with "xx" and the third occurrence of 'a' with "xxx". After the third occurrence, begin the replacement pattern over again with "x", "xx", "xxx"...etc.; however, if an 'a' is followed by more than 2 other 'a' characters in a row, then do not replace any more 'a' characters after that 'a'.
No use of the replace method is allowed.
aTo123X("ababba") → "xbxxbbxxx"
aTo123X("anaceeacdabnanbag") → "xnxxceexxxcdxbnxxnbxxxg"
aTo123X("aabaaaavfaajaaj") → "xxxbxxxaaavfaajaaj"
aTo123X("pakaaajaaaamnbaa") → "pxkxxxxxxjxxaaamnbaa"
aTo123X("aaaak") → "xaaak"
My code's output is with a's included, x's added but not the correct amount of x's.
public String aTo123X(String str) {
/*
Strategy:
get string length of the code, and create a for loop in order to find each individual part of the String chars.check for a values in string and take in pos of the a.
if one of the characters is a
replace with 1 x, however, there aren't more than 2 a's immediately following first a and as it keeps searching through the index, add more x's to the original string, but set x value back to 1 when x reaches 3.
if one of characters isn't a,
leave as is and continue string.
*/
String xVal = "";
String x = "x";
String output = "";
for (int i = 0; i < str.length(); i++){
if( str.charAt(i) == 'a'){
output += x;
str.substring(i+1, str.length());
}
output += str.charAt(i);
}
return output;
}
This is the code that does the same. I've commented the code to explain what it does
public class ReplaceChar {
public static void main(String... args){
String[] input =new String[]{"ababba","anaceeacdabnanbag","aabaaaavfaajaaj"};
StringBuilder result = new StringBuilder();
for (int i= 0; i < input.length;i++){
result.append(getReplacedA(input[i]));
result.append("\n");
}
System.out.println(result);
}
private static String getReplacedA(String withA){
// stringBuilder for result
StringBuilder replacedString = new StringBuilder();
// counting the number of time char 'a' occurred in String for replacement before row of 'aaa'
int charACount = 0;
// get the first index at which more than two 'aa' occurred in a row
int firstIndexOfAAA = withA.indexOf("aaa") + 1;
// if 'aaa' not occurred no need to add the rest substring
boolean addSubRequired = false;
// if the index is 0 continue till end
if (firstIndexOfAAA == 0)
firstIndexOfAAA = withA.length();
else
addSubRequired = true;
char[] charString = withA.toCharArray();
//Replace character String[] array
String[] replace = new String[]{"x","xx","xxx"};
for(int i = 0; i < firstIndexOfAAA; i++){
if (charString[i] == 'a'){
charACount++;
charACount = charACount > 3 ? 1 : charACount ;
// add the number x based on charCount
replacedString.append(replace[charACount - 1]);
}else{
replacedString.append(charString[i]);
}
}
// if the String 'aaa' has been found previously add the remaining subString
// after that index
if (addSubRequired)
replacedString.append(withA.substring(firstIndexOfAAA));
// return the result
return replacedString.toString();
}
}
Output:
xbxxbbxxx
xnxxceexxxcdxbnxxnbxxxg
xxxbxxxaaavfaajaaj
EDIT : Some Improvement You can make for some corner cases in the getReplacedA() function:
Check if char 'a' is there or not in the String if not just return the String No need to do anything further.
Use IgnoreCase to avoid the uppercase or lowercase possibility.
Firstly, string is immutable, so the below statement does nothing
str.substring(i+1, str.length());
I guess you wanted to do:
str = str.substring(i+1, str.length());
However, even after fix that, your program still doesn't work. I can't really comprehend your solution. 1) you are not detecting more than 3 a's in a row. 2) you are not appending "xx" or "xxx" at all
Here is my version, works for me so far:
public static void main(String[] args) {
System.out.println(aTo123X("ababba")); // "xbxxbbxxx"
System.out.println(aTo123X("anaceeacdabnanbag")); // "xnxxceexxxcdxbnxxnbxxxg"
System.out.println(aTo123X("aabaaaavfaajaaj")); // "xxxbxxxaaavfaajaaj"
}
public static String aTo123X(String str) {
String output = "";
int aOccurrence = 0;
String[] xs = {"x", "xx", "xxx"};
for (int i = 0; i < str.length(); ++i) {
if (str.charAt(i) == 'a') {
output += xs[aOccurrence % 3]; // append the x's depending on the number of a's we have seen, modulus 3 so that it forms a cycle of 3
if (i < str.length() - 3 && str.charAt(i + 1) == 'a' && str.charAt(i + 2) == 'a' && str.charAt(i + 3) == 'a') {//if an 'a' is followed by more than 2 other 'a' characters in a row
output += str.substring(i + 1);
break;
} else {
++aOccurrence; // increment the a's we have encountered so far
}
} else {
output += str.charAt(i); // append the character if it is not a
}
}
return output;
}
public class NewClass {
public static void main(String[] args) {
System.out.println(aTo123X("ababba")); // "xbxxbbxxx"
System.out.println(aTo123X("anaceeacdabnanbag")); // "xnxxceexxxcdxbnxxnbxxxg"
System.out.println(aTo123X("aabaaaavfaajaaj")); //xxxbxxxaaavfaajaaj
}
public static String aTo123X(String str) {
String output = "";
int aCount = 0;
int inRow = 0;
for (int i = 0; i < str.length();) {
if (str.charAt(i) == 'a') {
if (inRow <= 1) {
inRow++;
aCount++;
if (aCount == 1) {
output += "x";
} else if (aCount == 2) {
output += "xx";
} else {
output += "xxx";
aCount = 0;
}
boolean multiple = ((i + 1) < str.length()) && (str.charAt(i + 1) == 'a')
&& ((i + 2) < str.length()) && (str.charAt(i + 2) == 'a');
if (multiple) {
i++;
while (i < str.length()) {
output += str.charAt(i++);
}
return output;
}
} else {
output += str.charAt(i);
}
} else {
output += str.charAt(i);
inRow = 0;
}
i++;
}
return output;
}
}
I am pointing out problems in your code in form of comments in the code itself.
public String aTo123X(String str) {
//You are not using xVal variable in your code, hence it's obsolete
String xVal = "";
//You don't need x variable as you can simply use string concatenation
String x = "x";
String output = "";
for (int i = 0; i < str.length(); i++) {
/**
* Here, in "if" block you have not implmented any logic to replace the 2nd and
* 3rd occurence of 'a' with 'xx' and 'xxx' respectively. Also, substring() returns
* the sub-string of a string but you are not accepting that string anywhere, and
* you need not even use sub-string as "for" loop will cycle through all the
* characters in the string. If use sub-string method you code will only process
* alternative characters.
*/
if( str.charAt(i) == 'a') {
output += x;
str.substring(i+1, str.length());
}
/**
* Because of this statement a's are also returned, because this statement gets
* in both scenarios, whether the current character of string is a or not.
* But, this statement should get executed only when current character of the
* string is 'a'. So, in terms of coding this statement gets executed no matter
* "if" loop is executed or not, but it should get executed only when "if" loop
* is not executed. So, place this statement in else block.
*/
output += str.charAt(i);
}
return output;
}
I have implemented the logic for you. Here is Solution for your problem, just copy and run it. It passes all the specified test cases.
public String aTo123X(String str) {
String output = "";
int count = 1;
boolean flag = true;
for (int i = 0; i < str.length(); i++) {
if(str.charAt(i) == 'a' && flag == true) {
switch(count) {
case 1: output += "x";
count++;
break;
case 2: output += "xx";
count++;
break;
case 3: output += "xxx";
count = 1;
break;
}
if ((str.charAt(i+1) == 'a' && str.charAt(i+2) == 'a') == true) {
flag = false;
}
}
else {
output += str.charAt(i);
}
}
return output;
}
I use Map To store where to replace
public static void main(String[] args) {
System.out.println(aTo123X("ababba"));//xbxxbbxxx
System.out.println(aTo123X("anaceeacdabnanbag"));//xnxxceexxxcdxbnxxnbxxxg
System.out.println(aTo123X("aabaaaavfaajaaj"));//xxxbxxxaaavfaajaaj
}
public static String aTo123X(String str){
String res = "";
int nthReplace = 1; //Integer to store the nth occurence to replace
//Map to store [key == position of 'a' to replace]
//[value == x or xx or xxx]
Map<Integer, String> toReplacePos = new HashMap<>();
//The loop to know which 'a' to replace
for (int i = 0; i < str.length(); i++) {
if(str.charAt(i) == 'a'){
toReplacePos.put(i, nthReplace % 3 == 1 ? "x": (nthReplace % 3 == 2 ? "xx": "xxx"));
nthReplace++;
//Break if an 'a' is followed by more than 2 other 'a'
try {
if((str.charAt(i+1) == 'a')
&& (str.charAt(i+2) == 'a')
&& (str.charAt(i+3) == 'a')){
break;
}
} catch (StringIndexOutOfBoundsException e) {
}
}
}
//Do the replace
for (int i = 0; i < str.length(); i++) {
res += toReplacePos.containsKey(i) ? toReplacePos.get(i) : str.charAt(i);
}
return res;
}
I have edited my answer. This one is giving the correct solution:
public static void main (String[] args) throws InterruptedException, IOException, JSONException {
System.out.println(aTo123X("ababba")); //xbxxbbxxx
System.out.println(aTo123X("anaceeacdabnanbag")); //xnxxceexxxcdxbnxxnbxxxg
System.out.println(aTo123X("aabaaaavfaajaaj")); //xxxbxxxaaavfaajaaj
}
public static String aTo123X(String str) {
String x = "x";
String xx = "xx";
String xxx = "xxx";
int a = 1;
int brek = 0;
String output = "";
for (int i = 0; i < str.length(); i++) {
if(str.charAt(i) == 'a' && a == 1) {
output += x;
str.substring(i+1, str.length());
a = 2;
try {
if(str.charAt(i+1) == 'a' && str.charAt(i+2) == 'a')
brek += 1;
} catch (Exception e) {
// TODO Auto-generated catch block
e.printStackTrace();
}
}
else if(str.charAt(i) == 'a' && a == 2) {
output += xx;
str.substring(i+1, str.length());
a = 3;
try {
if(str.charAt(i+1) == 'a' && str.charAt(i+2) == 'a')
brek += 1;
} catch (Exception e) {
// TODO Auto-generated catch block
e.printStackTrace();
}
}
else if(str.charAt(i) == 'a' && a == 3) {
output += xxx;
str.substring(i+1, str.length());
a = 1;
try {
if(str.charAt(i+1) == 'a' && str.charAt(i+2) == 'a')
brek += 1;
} catch (Exception e) {
// TODO Auto-generated catch block
e.printStackTrace();
}
}
else {
output += str.charAt(i);
brek = 0;
}
if(brek>0) {
output += str.substring(i+1);
break;
}
}
return output;
}
I am attempting to read this .txt file into my program (as an improvement over manual input) and i am having trouble converting my methods to accept the input txt file. i get a arrayindexoutofboundsexception on line "infix[--pos]='\0';"
class Functions {
void postfix(char infix[], char post[]) {
int position, und = 1;
int outposition = 0;
char topsymb = '+';
char symb;
Stack opstk = new Stack();
opstk.top = -1;
for (position = 0; (symb = infix[position]) != '\0'; position++) {
if (isoperand(symb))
post[outposition++] = symb;
else {
if (opstk.isempty() == 1)
und = 1;
else {
und = 0;
topsymb = opstk.pop();
}
while (und == 0 && precedence(topsymb, symb) == 1) {
post[outposition++] = topsymb;
if (opstk.isempty() == 1)
und = 1;
else {
und = 0;
topsymb = opstk.pop();
}
}// end while
if (und == 0)
opstk.push(topsymb);
if (und == 1 || (symb != ')'))
opstk.push(symb);
else
topsymb = opstk.pop();
}// end else
}// end for
while (opstk.isempty() == 0)
post[outposition++] = opstk.pop();
post[outposition] = '\0';
}// end postfix function
int precedence(char topsymb, char symb) {
/* check precedence and return 0 or 1 */
if (topsymb == '(')
return 0;
if (symb == '(')
return 0;
if (symb == ')')
return 1;
if (topsymb == '$' && symb == '$')
return 0;
if (topsymb == '$' && symb != '$')
return 1;
if (topsymb != '$' && symb == '$')
return 0;
if ((topsymb == '*' || topsymb == '/') && (symb != '$'))
return 1;
if ((topsymb == '+' || topsymb == '-') && (symb == '-' || symb == '+'))
return 1;
if ((topsymb == '+' || topsymb == '-') && (symb == '*' || symb == '/'))
return 0;
return 1;
} /* end precedence function */
private boolean isoperand(char symb) {
/* Return 1 if symbol is digit and 0 otherwise */
if (symb >= '0' && symb <= '9')
return true;
else
return false;
}/* end isoperand function */
}
public class Driver {
public static void main(String[] args) throws IOException {
Functions f = new Functions();
char infix[] = new char[80];
char post[] = new char[80];
int pos = 0;
char c;
System.out.println("\nEnter an expression is infix form : ");
try {
BufferedReader in = new BufferedReader(new FileReader("infix.txt"));
String str;
while ((str = in.readLine()) != null) {
infix = str.toCharArray();
}
in.close();
} catch (IOException e) {
}
infix[--pos] = '\0';
System.out.println("The original infix expression is : ");
for (int i = 0; i < pos; i++)
System.out.print(infix[i]);
f.postfix(infix, post);
System.out.println("\nThe postfix expression is : ");
for (int i = 0; post[i] != '\0'; i++)
System.out.println(post[i]);
}
}
Do should never ever do like this:
try {
...
} catch (IOException e) {
}
You loose some essential information about your code-running.
At lease you should print the stack trace to follow the investigation:
e.printStackTrace();
You may have a FileNotFound exception.
In addition you try to index your array to -1 in infix[--pos], pos is set to 0 before this statement.
1) Totally aside, but I think line in main should read:System.out.println("\nEnter an expression in infix form : ");
2) As well, i agree about the catch statement. You already have narrowed it down to being an IOExcpetion, but you can find so much more info out by printing wither of the following inside the catch
System.err.println(e.getMessage()); or e.printStackTrace()
Now to answer your question. You are initializing pos to the value 0, but the you are doing a PREINCREMENT on the line infix[--pos] = '\0'; so pos becomes -1 (clearly outside the scope of the array bounddaries).
I think you want to change that to a post increment infix[pos--] = '\0';. Perhaps?
And yes, your code DOES Look like C...