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Oracle

1Z0-819 · Question #187

What change will cause the code to compile successfully? Given: public class Electronics { public Electronics(double price) { super(); } } and public class Plushy extends Electronics { public…

The correct answer is A. Insert `PriceChecker (?) prod` on line 1. Option A uses PriceChecker<?> (the wildcard type), which declares a single prod variable that can hold either PriceChecker<Electronics> or PriceChecker<Plushy>. This resolves the duplicate variable name - instead of two separate declarations, you declare prod once with the…

Question

What change will cause the code to compile successfully? Given: public class Electronics { public Electronics(double price) { super(); } } and public class Plushy extends Electronics { public Plushy(double price) { super(price); } } and public class PriceChecker <T extends Electronics> { private T product; public PriceChecker (T product) { this.product = product; } public boolean isPriceEqual(double price) { return prod.product.getprice() == price; // line 1 } public static void main(String... args) { PriceChecker<Electronics> prod = new PriceChecker<>(new Electronics(1000.00)); PriceChecker<Plushy> prod = new PriceChecker<>(new Plushy(1.2)); // problem: variable prod already defined System.out.println(prod.isPriceEqual(b)); // problem: b not defined. } }

Options

  • AInsert PriceChecker (?) prod on line 1.
  • BInsert PriceChecker <Electronics> prod on line 1.
  • CInsert PriceChecker <Plushy> prod on line 1.
  • DInsert PriceChecker <Object> prod on line 1.

How the community answered

(45 responses)
  • A
    84% (38)
  • B
    2% (1)
  • C
    9% (4)
  • D
    4% (2)

Explanation

Option A uses PriceChecker<?> (the wildcard type), which declares a single prod variable that can hold either PriceChecker<Electronics> or PriceChecker<Plushy>. This resolves the duplicate variable name - instead of two separate declarations, you declare prod once with the wildcard, then reassign it: the wildcard ? is compatible with any type argument that satisfies the class-level bound (T extends Electronics).

Why the others fail:

  • B (PriceChecker<Electronics>) won't work because Java generics are invariant - PriceChecker<Plushy> is not a subtype of PriceChecker<Electronics>, so the reassignment would cause a compile error.
  • C (PriceChecker<Plushy>) fails for the same invariance reason in reverse - PriceChecker<Electronics> is not a subtype of PriceChecker<Plushy>.
  • D (PriceChecker<Object>) violates the class's generic bound <T extends Electronics> - Object doesn't extend Electronics, so the compiler rejects it outright.

Memory tip: Unlike arrays, Java generics are invariant - List<Dog> is not a List<Animal>. When you need a single reference to hold multiple parameterized types, reach for the wildcard ?, which opts into the subtype flexibility that raw generics otherwise block.

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