Puzzle

Difficulty: Hard · Puzzle 4 of 6 · Topic: Clocking Block Puzzles

Two monitors sample the same grant with different input skews. Which one sees same-edge NBA-updated grant?

Code

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interface ifc(input bit clk);
  logic req, grant;
  clocking mon0 @(posedge clk);
    default input #0;
    input grant;
  endclocking
  clocking mon1 @(posedge clk);
    default input #1step;
    input grant;
  endclocking
endinterface

module tb;
  bit clk = 0;
  ifc vif(clk);
  always #5 clk = ~clk;

  initial vif.req = 1;
  always @(posedge clk) vif.grant <= vif.req;

  initial begin
    @(vif.mon0);
    $display("[%0t] mon0.grant=%0b mon1.grant=%0b", $time, vif.mon0.grant, vif.mon1.grant);
    $finish;
  end
endmodule

Hint

input #0 samples at event (after NBA visibility), input #1step samples before edge.

Step-by-step solution

1) grant<=req commits in NBA on the first edge.
2) mon0 input #0 samples value aligned with event and sees updated grant=1.
3) mon1 input #1step sampled just before edge and still sees old grant=0.

Answer

Answer: mon0 sees 1 while mon1 sees 0 on that first event. Skew choice directly changes sampled truth.

Why candidates get it wrong

Using #1step for 'safety' can accidentally introduce one-cycle stale sampling.

Interviewer follow-up

Which skew is better for scoreboard checks against flopped DUT outputs and why?

Related topics

• Back to Clocking Block Puzzles

• Concurrency & Timing Puzzles section