Bird
Raised Fist0
Verilogprogramming~10 mins

Comparator design in Verilog - Interactive Code Practice

Choose your learning style10 modes available
Practice - 5 Tasks
Answer the questions below
1fill in blank
easy

Complete the code to declare a 1-bit input named 'a'.

Verilog
input [1];
Drag options to blanks, or click blank then click option'
Aout
Bb
Ca
Dclk
Attempts:
3 left
💡 Hint
Common Mistakes
Using output instead of input
Forgetting to declare the signal name
2fill in blank
medium

Complete the code to compare inputs 'a' and 'b' and assign the result to 'out'.

Verilog
assign out = (a [1] b);
Drag options to blanks, or click blank then click option'
A>
B<=
C==
D!=
Attempts:
3 left
💡 Hint
Common Mistakes
Using equality operator instead of greater than
Using less than operator by mistake
3fill in blank
hard

Fix the error in the always block sensitivity list to trigger on changes of 'a' or 'b'.

Verilog
always @([1]) begin
  if (a > b)
    out = 1;
  else
    out = 0;
end
Drag options to blanks, or click blank then click option'
Aposedge a or posedge b
Ba or b
Ca, b
Da & b
Attempts:
3 left
💡 Hint
Common Mistakes
Using bitwise AND '&' instead of 'or'
Using commas ',' which is invalid in sensitivity list
4fill in blank
hard

Fill both blanks to complete the comparator module header and port list.

Verilog
module comparator([1], [2], out);
  input a, b;
  output out;
endmodule
Drag options to blanks, or click blank then click option'
Aa
Bout
Cb
Dclk
Attempts:
3 left
💡 Hint
Common Mistakes
Including output 'out' in the input port positions
Adding unrelated signals like 'clk'
5fill in blank
hard

Fill all three blanks to complete the conditional assignment inside the always block.

Verilog
always @(a or b) begin
  if ([1] [2] [3])
    out = 1;
  else
    out = 0;
end
Drag options to blanks, or click blank then click option'
Aa
B>
Cb
D==
Attempts:
3 left
💡 Hint
Common Mistakes
Using equality '==' instead of greater than '>'
Swapping 'a' and 'b' in the condition

Practice

(1/5)
1.

What is the main purpose of a comparator in Verilog?

easy
A. To store data temporarily
B. To check if one number is greater, less, or equal to another
C. To add two numbers together
D. To generate clock signals

Solution

  1. Step 1: Understand comparator function

    A comparator compares two values to determine their relationship.
  2. Step 2: Identify correct purpose

    Only To check if one number is greater, less, or equal to another describes comparing numbers, which is the comparator's job.
  3. Final Answer:

    To check if one number is greater, less, or equal to another -> Option B
  4. Quick Check:

    Comparator = Compare numbers [OK]
Hint: Comparator compares values, not adds or stores [OK]
Common Mistakes:
  • Confusing comparator with adder
  • Thinking comparator stores data
  • Mixing comparator with clock generator
2.

Which of the following is the correct way to start an always block for a comparator that reacts to changes in inputs a and b?

easy
A. always @(a, b) begin
B. always @(posedge clk) begin
C. always @(a and b) begin
D. always @(a or b) begin

Solution

  1. Step 1: Understand sensitivity list syntax

    In Verilog, sensitivity list uses 'or' to list signals that trigger the block.
  2. Step 2: Check options

    always @(a or b) begin uses 'or' correctly. always @(a, b) begin uses comma which is not valid in Verilog sensitivity lists.
  3. Final Answer:

    always @(a or b) begin -> Option D
  4. Quick Check:

    Sensitivity list uses 'or' between signals [OK]
Hint: Use 'or' between signals in sensitivity list [OK]
Common Mistakes:
  • Using commas instead of 'or' in sensitivity list
  • Using 'and' instead of 'or'
  • Using clock edge when not needed
3.

Given the following Verilog code snippet, what will be the output signals gt, lt, and eq if a = 5 and b = 3?

module comp(input [3:0] a, b, output reg gt, lt, eq);
  always @(a or b) begin
    if (a > b) begin
      gt = 1; lt = 0; eq = 0;
    end else if (a < b) begin
      gt = 0; lt = 1; eq = 0;
    end else begin
      gt = 0; lt = 0; eq = 1;
    end
  end
endmodule
medium
A. gt=0, lt=0, eq=1
B. gt=0, lt=1, eq=0
C. gt=1, lt=0, eq=0
D. gt=1, lt=1, eq=0

Solution

  1. Step 1: Compare values a=5 and b=3

    Since 5 is greater than 3, the condition 'a > b' is true.
  2. Step 2: Determine output signals

    For 'a > b', gt=1, lt=0, eq=0 as per code.
  3. Final Answer:

    gt=1, lt=0, eq=0 -> Option C
  4. Quick Check:

    5 > 3 means gt=1 [OK]
Hint: Check which condition is true first [OK]
Common Mistakes:
  • Mixing up greater than and less than
  • Setting multiple outputs to 1 incorrectly
  • Ignoring else conditions
4.

Identify the error in this comparator code snippet:

module comp(input [3:0] a, b, output reg gt, lt, eq);
  always @(a, b) begin
    if (a > b)
      gt = 1; lt = 0; eq = 0;
    else if (a < b)
      gt = 0; lt = 1; eq = 0;
    else
      gt = 0; lt = 0; eq = 1;
  end
endmodule
medium
A. Missing begin-end blocks after if and else if
B. Incorrect sensitivity list syntax
C. Output signals should be wires, not regs
D. Using non-blocking assignments instead of blocking

Solution

  1. Step 1: Check if-else statement structure

    Multiple statements after if need to be grouped with begin-end blocks.
  2. Step 2: Identify missing begin-end

    Here, multiple assignments after if and else if lack begin-end, causing syntax errors.
  3. Final Answer:

    Missing begin-end blocks after if and else if -> Option A
  4. Quick Check:

    Multiple statements need begin-end [OK]
Hint: Use begin-end for multiple statements in if blocks [OK]
Common Mistakes:
  • Forgetting begin-end with multiple statements
  • Confusing sensitivity list syntax
  • Wrong data type for outputs
5.

You want to design a 4-bit comparator that sets gt high if a > b, lt high if a < b, and eq high if equal. However, you also want to ignore the comparison if either a or b is zero (all bits zero), setting all outputs low in that case. Which Verilog code snippet correctly implements this?

hard
A. always @(a or b) begin if (a == 0 || b == 0) begin gt = 0; lt = 0; eq = 0; end else if (a > b) begin gt = 1; lt = 0; eq = 0; end else if (a < b) begin gt = 0; lt = 1; eq = 0; end else begin gt = 0; lt = 0; eq = 1; end end
B. always @(a or b) begin if (a == 0 && b == 0) begin gt = 0; lt = 0; eq = 0; end else if (a > b) begin gt = 1; lt = 0; eq = 0; end else if (a < b) begin gt = 0; lt = 1; eq = 0; end else begin gt = 0; lt = 0; eq = 1; end end
C. always @(a or b) begin if (a != 0 && b != 0) begin if (a > b) begin gt = 1; lt = 0; eq = 0; end else if (a < b) begin gt = 0; lt = 1; eq = 0; end else begin gt = 0; lt = 0; eq = 1; end end else begin gt = 0; lt = 0; eq = 0; end end
D. always @(a or b) begin if (a == 0 || b == 0) begin gt = 1; lt = 1; eq = 1; end else if (a > b) begin gt = 1; lt = 0; eq = 0; end else if (a < b) begin gt = 0; lt = 1; eq = 0; end else begin gt = 0; lt = 0; eq = 1; end end

Solution

  1. Step 1: Understand the zero check condition

    The requirement is to ignore comparison if either a or b is zero, so use 'a == 0 || b == 0'.
  2. Step 2: Verify output assignments

    always @(a or b) begin if (a == 0 || b == 0) begin gt = 0; lt = 0; eq = 0; end else if (a > b) begin gt = 1; lt = 0; eq = 0; end else if (a < b) begin gt = 0; lt = 1; eq = 0; end else begin gt = 0; lt = 0; eq = 1; end end sets all outputs low when either input is zero, then compares normally otherwise.
  3. Final Answer:

    Option A code snippet -> Option A
  4. Quick Check:

    Use '||' to check either zero, outputs low [OK]
Hint: Use '||' to check either zero, set outputs low [OK]
Common Mistakes:
  • Using '&&' instead of '||' for zero check
  • Setting outputs high when inputs are zero
  • Not handling zero inputs separately
Recommended

Ready to prove it?

Pass the test and this pattern moves unknown strong while earning +1 Grip toward your certificate.

10 questions~5 mineasy hard