Given: ``java package test.t1; public class A { public int x = 42; protected A() { } } package test.t2; import test.t1.; public class B extends A { int x = 17; public B() { super(); } } package…

Question

package test.t1;

public class A {
 public int x = 42;
 protected A() { }
}

package test.t2;

import test.t1.*;

public class B extends A {
 int x = 17;
 public B() { super(); }
}

package test;

import test.t1.*;

public class Tester {
 public static void main(String[] args) {
 A obj = new B(); // line 4
 System.out.println(obj.x); // line 5
 }
}

Options

• A42
• BThe compilation fails due to an error in line 4.
• C17
• DThe compilation fails due to an error in line 3.

• A
  72% (39)
• B
  4% (2)
• C
  7% (4)
• D
  17% (9)

Explanation

Option A (42) is correct because in Java, field access is statically bound - the compiler resolves obj.x based on the declared type of the reference (A), not the runtime type (B). Even though the object at runtime is a B instance (with x = 17), obj is declared as A, so obj.x reads A's field, which is 42. Option C (17) is the classic trap: unlike method calls, fields are not polymorphically dispatched - B.x simply hides A.x rather than overriding it, so the hidden field is only accessible through a reference typed as B. Options B and D are wrong because the code compiles cleanly - A obj = new B() is a valid upcast (B extends A), and B's constructor can legally call super() to invoke A's protected constructor since B is a subclass.

Memory tip: "Methods override, fields hide." Override = runtime type wins (dynamic dispatch). Hide = declared type wins (static resolution). Whenever you see a field accessed through an upcast reference on an exam, always use the declared type to find the value.

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