Given the method below, what is the widest valid return type for methodA on line 3? public class ReturnIt { returnType methodA(byte x, double y) // Line 3 { return (long)x / y * 2; } }

Difficulty: Easy

Correct Answer: double

Explanation:


Introduction / Context:
Expression type promotion in Java determines the resulting type of arithmetic expressions. Choosing a compatible return type requires analyzing how primitives are promoted.



Given Data / Assumptions:

  • x is a byte, explicitly cast to long.
  • y is a double.
  • The expression is (long)x / y * 2.


Concept / Approach:
Division involving a double promotes the other operand to double, and the result is double. Multiplying a double by an int literal 2 yields a double. Therefore, the expression overall is of type double, and the return type must be compatible (double or wider in the sense of assignment compatibility; here the precise primitive type is double).



Step-by-Step Solution:

Cast: (long)x → long.Operation: long / double → double.Operation: double * 2 → double.Conclusion: method must return double.


Verification / Alternative check:
Changing return type to double compiles; using long or int requires an explicit cast (and loses precision).



Why Other Options Are Wrong:
int, byte, and long are narrower than double for this expression; Number is not a primitive and does not match the expression type without boxing and is not a valid primitive return type here.



Common Pitfalls:
Assuming the initial cast to long dictates the overall type; forgetting that any double operand dominates.



Final Answer:
double

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