Last NBA in a Process Wins

Puzzle

Difficulty: Easy · Puzzle 5 of 6 · Topic: Event Scheduling Puzzles

Predict active print and end-of-slot value of q on the first edge. Explain why only one NBA survives.

Code

module tb;
  bit clk = 0;
  int q = 0;
  always #5 clk = ~clk;

  always @(posedge clk) begin
    q <= 1;
    q <= 2;
    q <= 3;
    $display("[%0t ACTIVE] q=%0d", $time, q);
  end

  initial begin
    @(posedge clk);
    $strobe("[%0t POSTPONED] q=%0d", $time, q);
    $finish;
  end
endmodule

Hint

Within one process and one time slot, later NBA to same variable overrides earlier NBA updates.

Step-by-step solution

1) Active print sees old q=0 at t=5.
2) Three NBAs are queued for q in same process.
3) Last assignment q<=3 overrides earlier q<=1 and q<=2 for that slot.
4) Postponed print shows q=3 at t=5.

Answer

Answer: Active shows q=0 at t=5; final value is q=3 at t=5. Last NBA to same variable in the process wins.

Why candidates get it wrong

People mix this with cross-process NBA ordering, which is not guaranteed.

Interviewer follow-up

What if each q<= assignment were in separate always blocks on the same edge?

Related topics

• Back to Event Scheduling Puzzles

• Concurrency & Timing Puzzles section