Given the code fragment: String shirts[][] = new String[2][2]; shirts[0][0] = "red"; shirts[0][1] = "blue"; shirts[1][0] = "small"; shirts[1][1] = "medium"; Which code fragment prints red: blue…

Question

Options

• Afor (int index = 1; index < 2; index++) { for (int idx = 1; idx < 2; idx++) { System.out.print (shirts[index][idx] + ":"); } }
• Bfor (int index = 0; index < 2; ++index) { for (int idx = 0; idx < index; ++idx) { System.out.print (shirts[index][idx] + ":"); } }
• Cfor (String c : colors) { for (String s : sizes) { System.out.println(s + ":"); } }
• Dfor (int index = 0; index < 2;) { for (int idx = 0; idx < 2;) { System.out.print (shirts[index][idx] + ":"); } idx++; }

• A
  7% (2)
• B
  4% (1)
• C
  18% (5)
• D
  71% (20)

Explanation

Option D is the intended correct answer because it initializes both loop counters to 0 and iterates through all four elements of the 2D array in order - shirts[0][0], shirts[0][1], shirts[1][0], shirts[1][1] - producing red: blue: small: medium:.

Why the distractors fail:

• A starts both index and idx at 1 instead of 0, skipping row 0 entirely and only printing medium:.
• B uses the condition idx < index in the inner loop; when index = 0, the inner loop body never executes, so only small: is printed on the second outer iteration.
• C references colors and sizes, which are never declared - this won't even compile, and it uses println which adds a newline rather than : as a separator.

Memory tip: For 2D array traversal, remember the two rules: both counters must start at 0 (not 1), and the inner loop's upper bound must be independent of the outer counter (not idx < index). When you see a loop starting at 1 or a nested bound tied to the outer variable, those are red flags that rows or columns will be skipped.

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