Given: ``java public class Test { public static void main(String[] args) { char[] arr = {97, 't', 'e', '\n', 'i', '\t', 'o'}; for (char var: arr) { System.out.print(var); } System.out.print("\nThe…

Question

public class Test {
 public static void main(String[] args) {
 char[] arr = {97, 't', 'e', '\n', 'i', '\t', 'o'};
 for (char var: arr) {
 System.out.print(var);
 }
 System.out.print("\nThe length is :" + arr.length);
 }
}

Options

• Aa e The length is : 2
• Ba e io The length is : 4
• Caeio The length is : 4
• Da e io The length is : 7
• ECompilation fails.

• A
  4% (1)
• B
  13% (3)
• C
  4% (1)
• D
  71% (17)
• E
  8% (2)

Explanation

Option D is correct because the char array has 7 elements, making arr.length equal to 7 - immediately eliminating A (length 2), B, and C (both length 4). The integer literal 97 is a compile-time constant that fits within the valid char range (0–65535), so Java silently promotes it to the character 'a' (ASCII 97), which is why option E is wrong - the code compiles cleanly. The escape sequences '\n' and '\t' are treated as single characters (newline and tab respectively), causing the output to split across two lines with i followed by a tab and then o on the second line; the final System.out.print starts with "\n", which pushes the length message to a third line.

Memory tip: For char arrays, remember "NAIL the length" - Newline (\n), ASCII numerics, Integer-to-char promotion, and Length counts all elements including invisible ones like \n and \t. Every element in the array counts toward length, even whitespace characters.

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