Given: ``java public class Test2 { public static void doChange(int[] arr) { for (int pos = 0; pos < arr.length; pos++){ arr[pos] = arr[pos] + 1; } } public static void main(String[] args) { int[]…

Question

public class Test2 {
 public static void doChange(int[] arr) {
 for (int pos = 0; pos < arr.length; pos++){
 arr[pos] = arr[pos] + 1;
 }
 }
 public static void main(String[] args) {
 int[] arr = {10, 20, 30};
 doChange(arr);
 for (int x : arr) {
 System.out.print(x + ", ");
 }
 
 doChange(arr[0], arr[1], arr[2]);
 System.out.print(arr[0] + ", " + arr[1] + ", " + arr[2]);
 }
}

Options

• A11, 21, 31, 11, 21, 31
• B11, 21, 31, 12, 22, 32
• C12, 22, 32, 12, 22, 32
• D10, 20, 30, 10, 20, 30
• ECompilation fails

• A
  3% (1)
• B
  13% (4)
• C
  7% (2)
• E
  77% (23)

Explanation

Compilation fails because doChange(arr[0], arr[1], arr[2]) attempts to call a method that expects a single int[] argument with three separate int values - no such overload exists, so the compiler rejects the code before it ever runs. This eliminates A, B, C, and D entirely: none of those outputs are reachable because the program never executes. A and D would require the second call to either leave values unchanged or increment them, B would require it to increment a second time, and C would require both calls to succeed - all impossible when compilation itself fails.

Memory tip: Individual array elements like arr[0] are just primitives (int), not arrays - passing them where an int[] is expected is a type mismatch the compiler catches immediately. When you see a method called with elements instead of the array itself, always check the method signature first.

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