将字符串转换为int数组时，为什么会得到空指针？

Question

My main method:

public static void main(String[] args) {
    Scanner input = new Scanner(System.in);
    String string1;
    string1 = input.next();

    LargeInteger firstInt = new LargeInteger(string1);

    System.out.printf("First integer: %s \n", firstInt.display());
}

LargeInteger class:

public class LargeInteger {

    private int[] intArray;

    //convert the strings to array
    public LargeInteger(String s) {
        for (int i = 0; i < s.length(); i++) {
            intArray[i] = Character.digit(s.charAt(i), 10); // in base 10
        }
    }

    //display the strings
    public String display() {
        String result = "";

        for (int i = 0; i < intArray.length; i++) {
            result += intArray[i];
        }
        return result.toString();
    }
}

Answer 1

You did not instantiate your array. You need something like:

   private int[] intArray = new int[SIZE];

where size is the length of your array.

Answer 2

private int[] intArray;

Member variables are null by default, so you need to initialize this.

Most likely you want it the same size as your string:

public LargeInteger(String s) {
    intArray = new int[s.length()]; // Create the actual array before you try to put anything in it
    for (int i = 0; i < s.length(); i++) {
        intArray[i] = Character.digit(s.charAt(i), 10); // in base 10
    }
}

Or you should use a container that resizes itself, like ArrayList .

Answer 3

You are not initialize the array intArray, that way you are getting error, here is the complete program

import java.util.Scanner;

class  TestForNull {

    public static void main(String[] args) {
        Scanner input = new Scanner(System.in);
        String string1;
        string1 = input.next(); 

        LargeInteger firstInt = new LargeInteger(string1);

        System.out.printf ("First integer: %s \n", firstInt.display());
    }

}

and this is LargeInteger

public class LargeInteger {

    private int[] intArray;
    //convert the strings to array
    public LargeInteger(String s) {
        intArray = new int[s.length()];
        for (int i = 0; i < s.length(); i++) {
            intArray[i] = Character.digit(s.charAt(i), 10); // in base 10
        }
    }

    //display the strings
    public String display() {           
          String result="";

          for (int i = 0; i < intArray.length; i++) {     
            result += intArray[i];
          }
          return result.toString();
        }   
}

Answer 4

Diferent approach through Integer.parseInt Integer.parseInt("yourInt");

To achieve your goal:

String a = "12345667788" //sample
String b = "";
int [] vecInt = new int[a.length()];  // The lack of initialization was your mistake as the above stated
for(int i=0; i< a.length(); i++)
{
   b = a.substring(0,1);
   a= a.substring(1);
   vecInt[i] = Integer.parseInt(b);
}

Please be aware of Double, long have far higher range then Integer which might be enough in your case to avoid an array!

Answer 5

you forgot to initialize the array intArray

I would recommend to use a java.util.List

Answer 6

You forgot to initialize the array. You have written it in constructor and the variables declared in method or constructor needs to be initialize at the same time.

Note : Implementing your logic in Constructor is not recommended unless and until you dont have any other choice.