Complete the Arrays as objects exercises before reviewing the solutions.

Review the Arrays as objects exercise 1 solution with AP CS Tutor Brandon Horn.

Original code

public static void main(String[] args)
{
    int[] nums = new int[] {1, 2, 3};

    System.out.println(Arrays.toString(nums));
    // prints: [1, 2, 3]

    mystery1(nums);
    
    System.out.println(Arrays.toString(nums));
}

public static void mystery1(int[] arr)
{
    arr[0] = 4;
}

Output

[1, 2, 3]
[4, 2, 3]

Explanation

All arguments in Java are passed by value. In the main method at the call to mystery1, the value of nums is the memory address of the array containing [1, 2, 3]. The value of arr is set to a copy of the value of nums. Both arr and nums point to (store the memory address of) the same array.

The statement arr[0] = 4; changes the value at index 0 in the array to which arr points.

See the bottom of this page for similar code without a method call.

Step by step memory diagram

Step 1

public static void main(String[] args)
{
    int[] nums = new int[] {1, 2, 3};

    System.out.println(Arrays.toString(nums));
    // prints: [1, 2, 3]

    // additional code not yet run
}

Memory diagram after Step 1

All arrays in Java are objects, regardless of the type of the values inside. nums stores the memory address of the array (the arrow) just as it would for any other object. The values inside the array are primitive types and are stored directly in the array.

Output after Step 1

[1, 2, 3]

Step 2

public static void main(String[] args)
{
    int[] nums = new int[] {1, 2, 3};

    System.out.println(Arrays.toString(nums));
    // prints: [1, 2, 3]

    mystery1(nums);
    
    // additional code not yet run
}

public static void mystery1(int[] arr)
{
    // additional code not yet run
}

Memory diagram after Step 2

The call to mystery1 passes a copy of the value of nums, which is memory address of the array (the arrow). nums and arr point to (store the memory address of) the same array.

Output after Step 2

[1, 2, 3]

Step 3

public static void main(String[] args)
{
    int[] nums = new int[] {1, 2, 3};

    System.out.println(Arrays.toString(nums));
    // prints: [1, 2, 3]

    mystery1(nums);
    
    // additional code not yet run
}

public static void mystery1(int[] arr)
{
    arr[0] = 4;
}

Memory diagram after Step 3

The statement arr[0] = 4; changes the value at index 0 in the array to which arr points. nums points to the same array.

Output after Step 3

[1, 2, 3]

Step 4

public static void main(String[] args)
{
    int[] nums = new int[] {1, 2, 3};

    System.out.println(Arrays.toString(nums));
    // prints: [1, 2, 3]

    mystery1(nums);
    
    System.out.println(Arrays.toString(nums));
}

Memory diagram after Step 4

After mystery1 returns (ends), arr (the parameter) goes away. nums points to (stores the memory location of) the same array as before the call; however, the value at index 0 is now 4.

Output after Step 4

[1, 2, 3]
[4, 2, 3]

Similar code without method call

public static void main(String[] args)
{
    int[] nums = new int[] {1, 2, 3};

    System.out.println(Arrays.toString(nums));
    // prints: [1, 2, 3]

    int[] arr = nums;
    arr[0] = 4;
    
    System.out.println(Arrays.toString(nums));
    // prints: [4, 2, 3]
}

This code is nearly the same as the original code.

The statement int[] arr = nums; creates a new variable arr that points to (stores the memory address of) the array to which nums points.

The memory diagrams are the same except that arr remains in scope until the main method ends.

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