How to fix “exception in thread main java.lang.stringindexoutofboundsexception string index out of range” errors

I'm trying to recreate the game of mastermind except a simplified version. I've been stuck on this one error and I don't know how to fix it.

String combo = "";
int plus = 0;
String in = "";
int right = 0;
int numbers = 6;
int length = 4;
int tries = 10;
int[] guessNums = new int[numbers];
int[] comboNums = new int[numbers];

rules();
combo = guesses();

for(int i = 0; i < length; i++)

{
if(in.charAt(i) == combo.charAt(i))
{
right++;
}
guessNums[in.charAt(i)-49]++;
comboNums[combo.charAt(i)-49]++;
}

for(int i = 0; i < numbers; i++)
{
while(comboNums[i] > 0 && guessNums[i] > 0)
{
plus++;
comboNums[i]–;
guessNums[i]–;
}
}
String reset = “\u001B[0m”;
System.out.print(“\t”);
String a = “” + right;
printRed(a);
System.out.println(" " + reset + “” + plus);
System.out.println(“\n”);

tries–;
return(right == length);


it says the error is in line if(in.charAt(i) == combo.charAt(i)).

#java #arrays #string #methods

What is GEEK

Buddha Community

Samuel Tucker

1548051505

In your code above, note that string in is equal to the empty string (“”) at first.

This fails when you run the following for loop because you are iterating i from 0 to length = 4. String in doesn’t have any characters in it, so you can’t do in.charAt(i) for any value of i.

for(int i = 0; i < length; i++) 
{
  if(in.charAt(i) == combo.charAt(i))
  {  
    right++;
  }  
  guessNums[in.charAt(i)-49]++;
  comboNums[combo.charAt(i)-49]++;
}

Tyrique  Littel

Tyrique Littel

1600135200

How to Install OpenJDK 11 on CentOS 8

What is OpenJDK?

OpenJDk or Open Java Development Kit is a free, open-source framework of the Java Platform, Standard Edition (or Java SE). It contains the virtual machine, the Java Class Library, and the Java compiler. The difference between the Oracle OpenJDK and Oracle JDK is that OpenJDK is a source code reference point for the open-source model. Simultaneously, the Oracle JDK is a continuation or advanced model of the OpenJDK, which is not open source and requires a license to use.

In this article, we will be installing OpenJDK on Centos 8.

#tutorials #alternatives #centos #centos 8 #configuration #dnf #frameworks #java #java development kit #java ee #java environment variables #java framework #java jdk #java jre #java platform #java sdk #java se #jdk #jre #open java development kit #open source #openjdk #openjdk 11 #openjdk 8 #openjdk runtime environment

How to fix “exception in thread main java.lang.stringindexoutofboundsexception string index out of range” errors

I'm trying to recreate the game of mastermind except a simplified version. I've been stuck on this one error and I don't know how to fix it.

String combo = "";
int plus = 0;
String in = "";
int right = 0;
int numbers = 6;
int length = 4;
int tries = 10;
int[] guessNums = new int[numbers];
int[] comboNums = new int[numbers];

rules();
combo = guesses();

for(int i = 0; i < length; i++)

{
if(in.charAt(i) == combo.charAt(i))
{
right++;
}
guessNums[in.charAt(i)-49]++;
comboNums[combo.charAt(i)-49]++;
}

for(int i = 0; i < numbers; i++)
{
while(comboNums[i] > 0 && guessNums[i] > 0)
{
plus++;
comboNums[i]–;
guessNums[i]–;
}
}
String reset = “\u001B[0m”;
System.out.print(“\t”);
String a = “” + right;
printRed(a);
System.out.println(" " + reset + “” + plus);
System.out.println(“\n”);

tries–;
return(right == length);


it says the error is in line if(in.charAt(i) == combo.charAt(i)).

#java #arrays #string #methods

Samanta  Moore

Samanta Moore

1621096440

Functions for Strings in Java

In this tutorial, you will learn how to make better use of built-in functions for Strings in Java to program more quickly, effectively, and aesthetically.

What Is a String?

Firstly, of course, we have to initialize our string. What is a string used for?

  • You want to look at your string as a line, not as a mass of symbols.
  • If you have a long text, you want to work with the words, not the letters.
  • If you have lots of information, you need functions that solve questions as quickly as possible.

#java #tutorial #java strings #java tutorial for beginners #java string #string tutorial

Samanta  Moore

Samanta Moore

1620458875

Going Beyond Java 8: Local Variable Type Inference (var) - DZone Java

According to some surveys, such as JetBrains’s great survey, Java 8 is currently the most used version of Java, despite being a 2014 release.

What you are reading is one in a series of articles titled ‘Going beyond Java 8,’ inspired by the contents of my book, Java for Aliens. These articles will guide you step-by-step through the most important features introduced to the language, starting from version 9. The aim is to make you aware of how important it is to move forward from Java 8, explaining the enormous advantages that the latest versions of the language offer.

In this article, we will talk about the most important new feature introduced with Java 10. Officially called local variable type inference, this feature is better known as the **introduction of the word **var. Despite the complicated name, it is actually quite a simple feature to use. However, some observations need to be made before we can see the impact that the introduction of the word var has on other pre-existing characteristics.

#java #java 11 #java 10 #java 12 #var #java 14 #java 13 #java 15 #verbosity

Macey  Legros

Macey Legros

1600530180

Java String Class Example | String Class in Java Tutorial

Java String is an inbuilt class that deals with all that is written in double quotation marks, or other words string is a collection of characters. For example – “Java” is a String of length 4.  Every string is an instance of this class. They are immutable, and so once written, and their value cannot be changed.

See the following figure.

Java String Class Example

The string is a sequence of characters. In Java, objects of String are immutable, which means a constant and cannot be changed once created. The java.lang.String class provides a lot of functions to work on the string. With the help of these functions, we can perform the operations on string such as trimming, concatenating, converting, comparing, replacing strings, etc.

#Why Strings are immutable or final in Java

Strings are used as parameters for many Java classes e.g.

  1. For opening a network connection.
  2. For opening a database connection.
  3. For File handling.

In case if strings were not immutable, this would have jeopardized the security.

#java #java string class #java.lang.string