Java arrays and command-line arguments: trace the output when copying args into a 2D array reference. public class CommandArgsThree { public static void main(String[] args) { String[][] argCopy = new String[2][2]; int x; argCopy[0] = args; // first row refers to args array x = argCopy[0].length; for (int y = 0; y < x; y++) { System.out.print(" " + argCopy[0][y]); } } } // Invocation: > java CommandArgsThree 1 2 3

Introduction / Context: The snippet checks understanding of Java arrays, references, and command-line arguments. Assigning argCopy[0] = args makes the first row reference the same one-dimensional array that holds all command-line strings, so iterating over argCopy[0] effectively iterates over args.

Given Data / Assumptions:

• args = {"1", "2", "3"} from the command line.
• argCopy is a 2x2 jagged 2D array whose first row is replaced to reference args.
• Loop bounds: y from 0 to x - 1, where x = argCopy[0].length = args.length = 3.

Concept / Approach: Because argCopy[0] is aliased to args, argCopy[0][y] equals args[y]. The loop prints each element with a leading space, producing “ 1 2 3”. MCQ options ignore the leading spaces and focus on tokens 1, 2, 3.

Step-by-Step Solution:

Verification / Alternative check: Printing argCopy[0] using Arrays.toString would show the same elements since both references point to the same backing array.

Why Other Options Are Wrong:

• 0 0 / 0 0 0: Not related to String contents.
• 1 2: Truncated; the loop covers all three elements.
• An exception: All indices are valid.

Common Pitfalls: Assuming argCopy[0] = args copies elements; it only reassigns the reference.

Final Answer: 1 2 3