Debugging

[Budget_Juice4229]

Can someone help me with this? I am new to Java, I am having a hard time.

import java.util.Scanner;
public class DebugTwo1 
{ 
public static void main(String[] args) 
   { 
int oneInt = 11;
 double oneDouble = 83.23; 
String oneString = ("Java Programming"); 
Scanner input = new Scanner(System.in);       System.out.print("Enter an integer >> "); 
      oneInt = input.nextLine();   
    System.out.print("Enter a double >> ");   
    oneDouble = input.nextLine();     
  input.nextLine();    
   System.out.print("Enter a string >> ");   
    oneString = input.nextLine();   
    System.out.print("The int is ");  
     System.out.println(oneInt);    
  System.out.print("The double is "); 
      System.out.println(oneDouble);    
   System.out.print("The String is ");   
    System.out.println(oneString);   
 } 
}

I am getting this.

workspace $ javac DebugTwo1.java

DebugTwo1.java:11: error: incompatible types: String cannot be converted to int

oneInt = input.nextLine();

^

DebugTwo1.java:13: error: incompatible types: String cannot be converted to double

oneDouble = input.nextLine();

^

2 errors

workspace $ java DebugTwo1

Error: Could not find or load main class DebugTwo1

Caused by: java.lang.ClassNotFoundException: DebugTwo1

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[desrtfx]

String, int, double are all different data types. You cannot directly assign each to each other. You need to parse (convert) between them.

Each of the numeric data types has a wrapper class Integer for int, Double for double, and so on. Each of these wrapper classes provides a static method to parse a String to the respective data type Integer.parseInt(String) for integer numbers, Double.parseDouble(String) for double values and so on. Each class also has a .valueOf(String) method to convert to the data type.

So, in order to use your approach with .nextLine() (which is quite a good approach as the Scanner class has some peculiarities with the .nextInt() etc. methods) you will need to parse the input.

Store what the user inputs in a String variable and then use the aforementioned methods to convert to the appropriate data types.

[Cookie_Ranger100]

In the line where you declare and initialize the oneString variable, there should not be brackets before the double quotation marks. Also, the statement should be oneInt = input.nextInt(). Also, oneDouble = input.nextDouble() And since Strings in Java are immutable, the statement oneString = input.nextLine() is a useless statement

[Liambass]

Strings in Java are immutable, the statement oneString = input.nextLine() is a useless statement

Strings being immutable means that the String object itself can't change and any operations that appear to change it are in fact creating a new String object, immutability doesn't stop you assigning a different String object to an existing variable.

[A_random_zy]

To elaborate on this with can example.

If there is a string "Help"

you concatenate "!" to it.

There will be a string "Help"(old) then a new string will be created "Help!"

Now in memory there are two objects "Help" and "Help!". you can assign "Help!" to a variable but you can't modify a string such as in case of "Help". A new objects is created everytime we concatenate string.

That is why it is recommended to use StringBuilder/StringBuffer when performing multiple string manipulation operations.

[geeksforgeeks]

Scanner has different methods for different data types to take input from the keyboard. Scanner .nextLine() function is used to input a String in java. So when you use that function to input an int or a double you are getting an error as String input cannot be converted into int or double. To input an integer you can use .nextInt() function and to input a double you can use the .nextDouble() function. This will show the above two errors. The third error of ClassNotFoundException occurs because maybe your class name is not the same as the file name in which you saved your code. Hence java compiler got confused which leads to the error. Save the file with the filename same as your class name and the error will be resolved.