Skip to content
DenseChip
Learn
PracticeCode labInterview

DenseChip

Navigate

Menu

LearnPracticeCode labInterview

Tools

All toolsSV editorCode repoResume
DenseChip

Practice, courses, and chip-design tools in one focused workspace.

LearnPracticeCode labInterviewToolsLearning paths

© 2026 DenseChip

AboutContactPrivacyTerms
Skip to lesson

Course contents

  • How to Use This Puzzle Bank
  • SystemVerilog Language Puzzles25
  • Difficulty Guide
  • Full Topic Map
  • Interview-Day Tips
  • OOP & Testbench Puzzles25
  • Concurrency & Timing Puzzles25
  • Randomization & Constraint Puzzles25
  • SVA Assertion Puzzles25
  • Functional Coverage Puzzles25
    • Covergroup and Coverpoint Puzzles
      • Auto vs explicit bins with illegal transition context
      • Default bin denominator puzzle
      • Bin array math with mixed declarations
      • Transition bin sequence counting
      • illegal_bins vs ignore_bins accounting
      • Same traffic, different denominator
    • Cross Coverage Puzzles
    • Coverage Options Puzzles
    • Coverage Closure and Sampling Timing Puzzles
    • Functional Coverage Puzzles: Tricky Q&A
  • Logic & Aptitude Puzzles25
  • Scenario & Debug Puzzles25
  • Interview Prep: Puzzle Rounds5
  • DV Puzzle Cheatsheet
Loading lesson

‹Covergroup and Coverpoint Puzzles
  1. Topics
  2. DV Interview Puzzles
  3. Functional Coverage Puzzles
  4. Covergroup and Coverpoint Puzzles
  5. Default bin denominator puzzle

VLSI DV Interview Puzzles · All levels

Default bin denominator puzzle

How many bins are scored, and what is the coverpoint percentage after the five samples?

On this page

    Puzzle

    Difficulty: Easy · Puzzle 2 of 6 · Topic: Covergroup and Coverpoint Puzzles

    How many bins are scored, and what is the coverpoint percentage after the five samples?

    Code

    design.sv
    1
    2
    3
    4
    5
    6
    7
    8
    9
    10
    11
    12
    13
    14
    15
    16
    17
    covergroup cg with function sample(bit [2:0] opcode);
      cp_opcode: coverpoint opcode {
        bins read   = {0};
        bins write  = {1};
        bins atomic = {2,3};
        bins other  = default;
      }
    endgroup
    
    initial begin
      cg c = new();
      c.sample(0);
      c.sample(1);
      c.sample(4);
      c.sample(7);
      c.sample(3);
    end

    Hint

    A named default bin is a scored bin, not metadata.

    Step-by-step solution

    Diagram
    1) Scored bins are read, write, atomic, other -> 4 bins total.
    2) Samples hit read (0), write (1), atomic (3), and other (4 and 7).
    3) All 4 bins are hit at least once -> 4/4 = 100%.

    Answer

    Answer: There are 4 scored bins and coverage is 100% (4 hit of 4).

    Why candidates get it wrong

    Engineers often forget default contributes to denominator once it is explicitly declared as a named bin.

    Interviewer follow-up

    How would the denominator change if other were replaced by ignore_bins other = default?

    Related topics

    • Back to Covergroup and Coverpoint Puzzles

    • Functional Coverage Puzzles section

    Previous

    131 / 210

    Next: Bin array math with mixed declarations

    Up next

    Up nextBin array math with mixed declarations

    Sign in to save progress

    Practice this lessonReport an error

    See also

    • Transition bin sequence counting
    • illegal_bins vs ignore_bins accounting
    • Same traffic, different denominator