In Java command-line arguments, what is printed by this program when executed as shown?\n\npublic class Test \n{ \n public static void main (String[] args) \n {\n String foo = args[1]; \n String bar = args[2]; \n String baz = args[3]; \n System.out.println("baz = " + baz); // Line 8\n } \n}\n\nInvocation:\n> java Test red green blue

Introduction / Context:
This question tests Java command-line argument indexing and what happens when you access an array element that does not exist. The code prints the value of args[3] after assigning several variables from the args array. Understanding zero-based indexing and the number of arguments passed is crucial.

Given Data / Assumptions:

• Program is executed with: java Test red green blue
• Therefore: args[0] = "red", args[1] = "green", args[2] = "blue"
• The code reads args[1], args[2], args[3]

Concept / Approach:
Java arrays are zero-indexed. If the length is 3, valid indices are 0, 1, and 2. Any attempt to access args[3] when only three elements are present will throw ArrayIndexOutOfBoundsException at runtime. This occurs before the print statement executes, so no normal output line is printed.

Step-by-Step Solution:

Verification / Alternative check:
If the invocation had four tokens after the class name (e.g., java Test a b c d), then args[3] would be "d" and the print would succeed. The provided invocation has only three arguments.

Why Other Options Are Wrong:

• baz = or baz = null: No line prints because the exception occurs before printing.
• baz = blue: That would require args[3] to exist; it does not.
• No output is produced: While effectively true in terms of normal output, the precise outcome is a runtime exception, which is the most accurate choice.

Common Pitfalls:
Forgetting zero-based indexing and assuming the first argument is args[1]; or assuming Java silently ignores out-of-range accesses (it does not).

Final Answer:
Runtime Exception