how is this add method supposed to add 2 numbers if it only accepts 1 parameter?
<p>I am working on an assignment where I need to do calculations with large numbers without using bignumber class. </p>

<p>I understand the theory now on how I will be adding the 2 numbers ( i have them taken in as strings, then inserted into int arrays) I will add the last digits and take anything over 9 and add it to the next. </p>

<p>I have been supplied with this method to use though:</p>

<pre><code>LargeInteger sum = firstInt.add(secondInt);
</code></pre>

<p>I'm a bit confused as to how I can make this method in that it only takes 1 of the 2 numbers in the parameters. </p>

<p>Here is the rest of the pertinant code I have done so far:</p>

<p>main:</p>

<pre><code>public static void main(String[] args) {
    Scanner input = new Scanner(System.in);
    String string1;
    String string2;
    int exp =0;

    System.out.print("Enter the first integer: ");
    //Store up the input string “string1” entered by the user from the keyboard.
    string1 = input.next(); 

    LargeInteger firstInt = new LargeInteger(string1);

    System.out.print("Enter the second integer: ");
    string2 = input.next(); 
    //Store up the input string “string2” entered by the user from the keyboard.
    LargeInteger secondInt = new LargeInteger(string2);

    LargeInteger sum = firstInt.add(secondInt);

    System.out.printf (" Sum = %s \n", sum.display());

}
</code></pre>

<p>And here is my second class LargeInteger (without the add method yet):</p>

<pre><code>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 &lt; 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 &lt; intArray.length; i++) {     
            result += intArray[i];
        }
        return result.toString();
    }   
}
</code></pre>
