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 & b3 | b2 ) System.out.print("ok "); if ( b1 & b2 | b2 & b3 | b2 | b1 ) System.out.println("dokey"); } }

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:

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