Question 3 of 481
Logical Operator Precedence
Given: boolean a = false; boolean b = true; boolean c = true; System.out.println(a || b && c); What is the output?
Quick Answer
The answer is true. This is correct because in Java, logical operator precedence dictates that the AND operator (&&) is evaluated before the OR operator (||), so the expression `a || b && c` is interpreted as `a || (b && c)`. With `a` false, `b` true, and `c` true, the inner `b && c` yields true, and then `false || true` evaluates to true. On the Oracle Java Foundations 1Z0-811 exam, this concept tests your understanding of how Java groups logical conditions without explicit parentheses, a common source of errors where test-takers mistakenly evaluate left-to-right. A frequent trap is assuming OR has equal or higher precedence, leading to an incorrect false result. To remember the hierarchy, think of AND as multiplication and OR as addition in Boolean algebra—just as multiplication binds tighter than addition, so does && bind tighter than ||.
⚠ Common exam trap
The trap here is that candidates often evaluate the expression left-to-right without considering operator precedence, mistakenly thinking `a || b` is evaluated first (which would be `true`) and then `&& c` would produce `true && true` = `true`, but the actual precedence changes the grouping, though in this specific case both orders yield `true`; however, the trap is to test whether you know the precedence rule, not just the outcome.
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
✓
true
In Java, the logical AND operator (&&) has higher precedence than the logical OR operator (||). Therefore, the expression `a || b && c` is evaluated as `a || (b && c)`. Given `a = false`, `b = true`, and `c = true`, `b && c` evaluates to `true`, and then `false || true` evaluates to `true`. Thus, the output is `true`, making option D correct.
Answer analysis
Option-by-option breakdown
For each option: why learners choose it and why it is or isn't the right answer here.
- ✗
false
Why it's wrong here
Incorrect.
- ✗
Compilation fails
Why it's wrong here
Incorrect.
- ✗
None of the above
Why it's wrong here
Incorrect.
- ✓
true
Why this is correct
Correct due to operator precedence.
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Same concept, more angles
2 more ways this is tested on 1Z0-811
These questions test the same concept from different angles. Work through them to make sure you can recognise it however the exam phrases it.
Variation 1. A developer writes: boolean b = !true && false; What is the value of b?
easy- A.Compilation error: invalid operator.
- ✓ B.false
- C.NullPointerException
- D.true
Why B: The expression `!true && false` is evaluated as `(!true) && false`, which is `false && false`. The logical AND (`&&`) operator returns `true` only if both operands are `true`; otherwise, it returns `false`. Therefore, `b` is assigned `false`. Option B is correct.
Variation 2. Given boolean a = true, b = false, c = true; What is the result of (a || b) && (b || c)?
hard- A.Short-circuit evaluation prevents evaluation
- ✓ B.true
- C.Compilation error
- D.false
Why B: a||b = true, b||c = true, true && true = true.
Last reviewed: Jun 11, 2026
This 1Z0-811 practice question is part of Courseiva's free Oracle certification practice question bank. Courseiva provides original exam-style practice questions with explanations, topic-based practice, mock exams, readiness tracking, and study analytics to help learners prepare for the 1Z0-811 exam.
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