Puzzle

Difficulty: Medium · Puzzle 2 of 6 · Topic: Race Condition Puzzles

For each edge, determine the active print and the final post-NBA value of s.

Code

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module tb;
  bit clk = 0;
  int s = 9;
  always #5 clk = ~clk;

  always @(posedge clk) begin
    s = 1;
    $display("[%0t A ACTIVE] s=%0d", $time, s);
  end

  always @(posedge clk) begin
    s <= 0;
    $display("[%0t B ACTIVE] scheduled s<=0", $time);
  end

  initial begin
    repeat (2) begin
      @(posedge clk);
      $strobe("[%0t POST] s=%0d", $time, s);
    end
    $finish;
  end
endmodule

Hint

Blocking updates active state; NBA commits at end of slot.

Step-by-step solution

1) Active region may print A before B or B before A.
2) A always drives active s to 1 at some point in that region.
3) B always schedules s<=0 for NBA.
4) Postponed value after NBA is always s=0 each edge.

Answer

Answer: Active print order can vary, but final s is deterministically 0 at t=5 and t=15 due to NBA commit.

Why candidates get it wrong

Confusing unstable active snapshots with final end-of-slot state causes wrong conclusions.

Interviewer follow-up

If B used blocking s = 0 instead, would final value stay deterministic?

Related topics

• Back to Race Condition Puzzles

• Concurrency & Timing Puzzles section