Multiple choice technology programming languages

Given: class Knowing { static final long tooth = 343L; static long doIt(long tooth) { System.out.print(++tooth + " "); return ++tooth; } public static void main(String[] args) { System.out.print(tooth + " "); final long tooth = 340L; new Knowing().doIt(tooth); System.out.println(tooth); } } What is the result?

  1. 343 340 340

  2. 343 340 342

  3. 343 341 342

  4. 343 341 340

  5. 343 341 343

  6. Compilation fails.

Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

First, static tooth (343L) is printed. Then the local final tooth (340L) is declared. doIt() receives 340, ++tooth makes it 341 (printed), then ++tooth makes it 342 (returned). Back in main, the local tooth is still 341 (not affected by doIt's parameter modifications). Output: 343 341 340.

AI explanation

main() first prints the static field tooth (343), giving '343 '. Then a local final variable tooth=340 shadows the static field for the rest of main. Calling doIt(tooth) passes a copy of 340 into the method's own parameter named tooth (a separate variable). Inside doIt: ++tooth pre-increments the parameter to 341 and prints '341 '; the second ++tooth increments it to 342 and returns it, but the caller discards the return value. Back in main, the local final tooth is untouched by the method call (Java is pass-by-value, and the local was never reassigned) — println(tooth) prints 340. Concatenating the three prints yields '343 341 340', matching option D. No other option correctly tracks that doIt operates on its own copy of the variable.