Sum two arrays element-by-element in Java

One more answer, using streams and providing a more generic solution:

import org.junit.Assert;
import org.junit.Test;

import java.util.function.IntBinaryOperator;
import java.util.stream.IntStream;

public class SOTest {

    @Test
    public void test() {
        int[] a = {0, 1, 2};
        int[] b = {3, 4, 5};

        int[] sum = applyOn2Arrays((x, y) -> x + y, a, b);
        int[] diff = applyOn2Arrays((x, y) -> x - y, a, b);
        int[] mult = applyOn2Arrays((x, y) -> x * y, a, b);


        Assert.assertArrayEquals(new int [] {3,5,7}, sum);
        Assert.assertArrayEquals(new int [] {-3,-3,-3}, diff);
        Assert.assertArrayEquals(new int [] {0,4,10}, mult);
    }

    private int[] applyOn2Arrays(IntBinaryOperator operator, int[] a, int b[]) {
        return IntStream.range(0, a.length)
                .map(index -> operator.applyAsInt(a[index], b[index]))
                .toArray();
    }
}

There's certainly nothing to enable this in the language. I don't know of anything in the standard libraries either, but it's trivial to put the code you've written into a utility method which you can call from anywhere you need it.

check this one: used sum and carry

public class SumOfTwoArrays{    

    public static int[] sumOfTwoArrays(int[] arr1, int[] arr2){

        int na = arr1.length;
        int nb = arr2.length;
        int nc;
        int min;

        if(na > nb){
            nc = na + 1;
            min = nb;
        }else{
            nc = nb + 1;
            min = na;
        }

        int[] c = new int[nc];
        int sum = 0;
        int carry = 0;
        int i = na - 1;
        int j = nb - 1;
        int k = nc - 1;

        while(i >= 0 && j>=0){
            sum  = arr1[i] + arr2[j] + carry;
            i--;
            j--;
            c[k] = sum % 10;
            sum = sum / 10;
            carry = sum % 10;
            k--;        
        }//end of while loop

        while(i >= 0){          //n2 has exhausted
            sum  = arr1[i] + carry;
            c[k] = sum % 10;
            sum = sum / 10;
            carry = sum % 10;
            i--;
            k--;
        }

        while(j >= 0){        //n1 has exhausted  
            sum  = arr2[j] + carry;
            c[k] = sum % 10;
            sum = sum / 10;
            carry = sum % 10;
            k--;        
            j--;
        }

        c[k] = carry;

        return c;
    }
}

Input: arr1 = {2, 1, 3}; arr2 = {3, 2};

Output: c: {0, 2, 4 ,5}