Java programming — Boolean operators with single ampersand/pipe (non–short-circuit) and overall truth evaluation: class SSBool { public static void main(String [] args) { boolean b1 = true; boolean b2 = false; boolean b3 = true; if ( b1 & b2 | b2 &.

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Solve this multiple-choice question and choose the correct option.

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Introduction / Context:This question focuses on non–short-circuit boolean operators & and | and how they combine in expressions to produce a final truth value. Given Data / Assumptions: b1 = true, b2 = false, b3 = true. Operators used: & and | on booleans (both evaluate both operands). Concept / Approach:Compute each term and then OR the results. Remember & yields true only if both sides are true, and | yields true if either side is true. Step-by-Step Solution: First if: b1 & b2 = true & false = false; b2 & b3 = false & true = false; expression becomes false | false | false = false; nothing printed.Second if: same first parts are false, but there is an extra | b1 at the end; false | false | false | true = true, so prints "dokey". Verification / Alternative check:Replace with parentheses to verify grouping, or temporarily convert to booleans to visualize each term. Why Other Options Are Wrong:They either print "ok" incorrectly or claim no output/compile error, both of which contradict the truth table. Common Pitfalls:Confusing & and | with their short-circuit counterparts && and ||; mixing precedence without parentheses (though all terms here are associative with |). Final Answer:dokey

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