Trace Java if/else with a dangling semicolon and a default boolean: what value prints for hand? public class If1 { static boolean b; // default false public static void main(String [] args) { short hand = 42; if (hand < 50 && !b) // Line 7 hand++; if (hand > 50); // Line 9 (dangling semicolon) else if (hand > 40) { hand += 7; hand++; } else { --hand; } System.out.println(hand); } }

Introduction / Context:
This problem focuses on Java control flow: short-circuit logic, default initialization of static booleans, and especially the effect of a stray semicolon after an if (a “dangling semicolon”).

Given Data / Assumptions:

• b is a static boolean with default value false.
• Start: hand = 42.
• There is a semicolon right after if (hand > 50) on Line 9.

Concept / Approach:
Evaluate conditions stepwise. if (hand < 50 && !b) is true, so hand++ executes. The second if has a semicolon as its statement, so the subsequent else if pairs with that if but the then-part is a no-op. The else-if block can still run depending on the condition outcome.

Step-by-Step Solution:
Initially: hand = 42, b = false. Check hand < 50 && !b → 42 < 50 and !false → true → hand++ → hand = 43. Next: if (hand > 50); → condition is false but the statement is the semicolon (does nothing). The else if (hand > 40) now evaluates: 43 > 40 → true → execute block. Block does hand += 7 → 50, then hand++ → 51. Prints 51.

Verification / Alternative check:
Remove the semicolon on Line 9; then the else-if would be associated normally, producing a different flow depending on the updated value of hand.

Why Other Options Are Wrong:
41 or 42 ignore the first increment; 50 ignores the second increment; 49 cannot occur under the shown path.

Common Pitfalls:
Not noticing the stray semicolon; misunderstanding how else binds to the nearest unmatched if; ignoring default values of static primitives.

Final Answer:
51