1Z0-811 Primitives, Strings and Operators Practice Question
A developer is writing a bitmask validation method. The method should return true if both input integers (x and y) have exactly the same least significant bit set. The developer writes: if (x & y == 1) { return true; } However, the condition never evaluates to true even when both numbers are odd (least significant bit = 1). Debugging shows that x and y are positive integers. What is the root cause and the correct fix?
Answer choices
Why each option matters
Answer the question above first, then reveal the full breakdown to understand why each option is right or wrong.
Correct answer & explanation
✓
Add parentheses: if ((x & y) == 1) { ... }
In Java, the bitwise AND operator & has lower precedence than the equality operator ==. So x & y == 1 is parsed as x & (y == 1), which first evaluates y == 1 to a boolean, then performs bitwise AND with x, treating true as 1 and false as 0. That's not the intended check. The correct fix is to add parentheses to ensure & is evaluated first: (x & y) == 1. This checks if both numbers have the same least significant bit set (both odd). Option B correctly adds parentheses. Options A, C, and D do not correctly address the bitwise check: A uses modulo which works for odd/even but not general bitmasking; C uses bitwise OR which checks if either bit is set; D uses logical AND which is for booleans, not integers.
Answer analysis
Option-by-option breakdown
For each option: why learners choose it and why it is or isn't the right answer here.
- ✗
Use modulo: if (x % 2 == 1 && y % 2 == 1) { ... }
Why it's wrong here
Modulo works to check if both are odd, but the requirement is about bitmasking using the & operator; this changes the approach entirely.
- ✓
Add parentheses: if ((x & y) == 1) { ... }
Why this is correct
Parentheses ensure the bitwise AND is performed before the comparison, correctly checking if the least significant bit of x & y is 1.
- ✗
Use bitwise OR: if (x | y == 1) { ... }
Why it's wrong here
Bitwise OR does not check that both bits are set; it checks if at least one is set. Also, the same precedence issue applies.
- ✗
Use logical AND: if (x && y == 1) { ... }
Why it's wrong here
&& is a logical operator and requires boolean operands; x is an integer, so this will not compile.
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