Java booleans, wrapper Boolean, and the single ampersand operator: determine the output (JDK 1.6+).\n\npublic class BoolTest {\n public static void main(String[] args) {\n Boolean b1 = new Boolean("false");\n boolean b2;\n b2 = b1.booleanValue();\n if (!b2) {\n b2 = true;\n System.out.print("x ");\n }\n if (b1 & b2) { // single &: evaluates both sides\n System.out.print("y ");\n }\n System.out.println("z");\n }\n}

Introduction / Context:
The task probes differences between primitive boolean and wrapper Boolean, the effect of booleanValue(), and the semantics of the single ampersand operator (&) in boolean expressions, which evaluates both operands but still returns a boolean result.

Given Data / Assumptions:

• b1 = new Boolean("false") so b1.booleanValue() is false.
• First if uses !b2, which is true because b2 is false.
• Single & is a non-short-circuit boolean operator.
• Output tokens have trailing spaces as shown.

Concept / Approach:
After the first if block, b2 becomes true and "x " is printed. The second condition evaluates (b1 & b2) → (false & true) which is false, so the body is skipped. Finally, println prints "z" on the same line after a space from the previous print.

Step-by-Step Solution:

Verification / Alternative check:
Replacing & with && would short-circuit, but the result would still be false for (false && true), producing the same printed tokens.

Why Other Options Are Wrong:

• z: Misses the earlier "x ".
• y z / x y z: The condition for y is false.
• Compilation fails: All types and methods are valid.

Common Pitfalls:
Confusing & (non-short-circuit) with && (short-circuit); misunderstanding wrapper vs primitive behavior.

Final Answer:
x z