如何检查arraylist的int []类型的对象？

Question

I'm trying to find out particular elements in my ArrayList (arrayOne). Each element should be an int[] . I've tried System.out.println(arrayOne) , which compiles but gives a irregular and strange number "[[I@370968]".

I've also tried System.out.println(arrayOne[0]) but it won't compile and emits the error

Array required but java.util.ArrayList found.

Given is the following code, with {1,12,3,13,123,2} passed to eg :

import java.util.ArrayList;

public class arrayTest {
    private ArrayList<int[]> arrayOne; 

    public arrayTest(int[] eg) {
        int[] xy = new int[2];
        arrayOne = new ArrayList<int[]>(eg.length);
        for (int i = 0; i < eg.length; i++) {
            int sv = String.valueOf(eg[i]).length();
            if (sv == 1) {
                xy[0] = 0;
                xy[1] = eg[i];
                arrayOne.add(xy);
            }
            else if (sv == 2) {
                System.out.println("two digits");
                // TODO add code to make xy[0] = the first
                // digit of eg and xy[1] = the second digit
            }
            else {
                System.out.println("too many digits");
                // and throw error accordingly
            }
            System.out.println(arrayOne);
        }
    }
}

1. How do make sure and print out the int array at arrayOne[0]
2. Given the code above if (sv == 2) and i want to split each individual number into an int[] with [0] being the first digit and [1] being the second digit how would i get the int value of each individual digit.

Answer 1

Use Arrays.toString(yourArray); to print out arrays in human readable form.

Answer 2

First of all, the string [I@370968 is displayed because you are trying to print an int[] , which is actually an object. Because this object does not override the object's toString() method, that method is derived from the Object class. The Object.toString() implementation, which prints the class name (in this case [I , because it is an int array), then an @ sign, and then the hash code of the object.

Your ArrayList contains a number of int[] s. Because an ArrayList is not an array (the one with the square brackets, like int[] ), you can't call an element on it as if it were an array. In short, you cannot call arrayOne[someDesiredIndex] .

In order to get an element from the ArrayList, call get(int index) on it; it returns the desired int[] . As already pointed out by another answer, you can use Arrays.toString(int[]) to print it in a human readable form.

To answer your questions:

1. You can retrieve the first index ( 0 ) of the first array inside arrayOne with the following code: arrayOne.get(0)[0] .

2. The following code should work:

    private static int[] intToArray(int n) { String str = String.valueOf(n); int length = str.length(); int[] ints = new int[length]; for (int i = 0; i < length; i++) { ints[i] = Integer.parseInt(str.substring(i, i + 1)); } return ints; } 

   Above method puts each digit into the next array position (it also works with digits greater than 99). With this method you can easily get each individual digit:

    int[] digits = intToArray(47); int a = digits[0]; // Will be 4 int b = digits[1]; // Will be 7 

---

So this is the class rewritten:

public class Rewrite {

    private ArrayList<int[]> arrayOne = new ArrayList<int[]>();

    public Rewrite(int[] eg) {
        for (int i = 0; i < eg.length; i++) {
            int length = String.valueOf(eg[i]).length();
            switch (length) {
                case 1:
                    this.arrayOne.add(new int[] { 0, eg[i] });
                    break;
                case 2:
                    this.arrayOne.add(intToArray(eg[i]));
                    break;
                default:
                    throw new IllegalArgumentException("Number " + eg[i] + " has too many digits");
                    // Or display the error or something.
            }
            System.out.println(Arrays.toString(this.arrayOne.get(i)));
        }
    }

    private static int[] intToArray(int n) {
        String str = String.valueOf(n);
        int length = str.length();
        int[] ints = new int[length];
        for (int i = 0; i < length; i++) {
            ints[i] = Integer.parseInt(str.substring(i, i + 1));
        }
        return ints;
    }

    public static void main(String[] args) {
        Rewrite r = new Rewrite(new int[] { 47, 53, 91, 8 });
    }