When to use blocking (combinational) in Verilog - Interactive Code Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to assign the output as the AND of inputs using blocking assignment.

Verilog

always @(*) begin
  out = [1] & b;
end

Drag options to blanks, or click blank then click option'

Aa

Bout

Cc

Dd

Attempts:

3 left

💡 Hint

Common Mistakes

Using the output variable on the right side causes incorrect logic.

Using non-input variables in the expression.

Use blocking assignment for combinational logic. Here, a is the input to be ANDed with b.

2fill in blank

medium

Complete the code to use blocking assignment for combinational logic that sums two inputs.

Verilog

always @(*) begin
  sum = [1] + b;
end

Drag options to blanks, or click blank then click option'

Aa

Bc

Cd

Dsum

Attempts:

3 left

💡 Hint

Common Mistakes

Using the output variable on the right side.

Using a variable not declared as input.

Use blocking assignment with inputs on the right side. Here, a and b are inputs summed to produce sum.

3fill in blank

hard

Fix the error in the combinational block by choosing the correct blocking assignment variable.

Verilog

always @(*) begin
  [1] = a & b;
  c = [1] | d;
end

Drag options to blanks, or click blank then click option'

Ad

Bc

Ca

Dtemp

Attempts:

3 left

💡 Hint

Common Mistakes

Using output variable c on the left side before it is assigned.

Using input variables incorrectly.

Use a temporary variable temp with blocking assignment to hold intermediate combinational results.

4fill in blank

hard

Fill both blanks to create a combinational block that calculates sum and carry using blocking assignments.

Verilog

always @(*) begin
  [1] = a ^ b;
  [2] = a & b;
end

Drag options to blanks, or click blank then click option'

Asum

Bcarry

Ctemp

Dresult

Attempts:

3 left

💡 Hint

Common Mistakes

Mixing up sum and carry variables.

Using non-blocking assignments in combinational logic.

Use blocking assignments to calculate sum as XOR and carry as AND of inputs a and b.

5fill in blank

hard

Fill all three blanks to write a combinational block that computes max, min, and mid values using blocking assignments.

Verilog

always @(*) begin
  [1] = (a > b) ? a : b;
  [2] = (a < b) ? a : b;
  [3] = (a != b) ? ([1] + [2]) >> 1 : a;
end

Drag options to blanks, or click blank then click option'

Amax_val

Bmin_val

Cmid_val

Dtemp

Attempts:

3 left

💡 Hint

Common Mistakes

Using non-blocking assignments in combinational blocks.

Mixing variable names or forgetting to assign all outputs.

Use blocking assignments to assign max_val, min_val, and mid_val in combinational logic.

Practice

1. Which type of assignment should you use inside an always @(*) block to model combinational logic in Verilog?

easy

A. Continuous assignment (assign)

B. Non-blocking assignment (<=)

C. Blocking assignment (=)

D. Initial assignment

Solution

1. Step 1: Understand the purpose of always @(*) blocks

   These blocks are used to describe combinational logic where outputs depend immediately on inputs.

2. Step 2: Choose the correct assignment type for combinational logic

   Blocking assignments (=) execute statements sequentially and update values immediately, which is suitable for combinational logic.

3. Final Answer:

   Blocking assignment (=) -> Option C

4. Quick Check:

   Combinational logic uses blocking (=) [OK]

Hint: Use '=' in always @(*) for immediate updates [OK]

Common Mistakes:

• Using non-blocking (<=) in combinational blocks
• Confusing continuous assign with blocking inside always
• Using initial blocks for combinational logic

2. Which of the following is the correct syntax for a blocking assignment inside an always @(*) block?

easy

A. output <= input + 1;

B. output = input + 1;

C. assign output = input + 1;

D. initial output = input + 1;

Solution

1. Step 1: Identify the assignment type inside always @(*)

   Blocking assignments use the '=' operator inside procedural blocks.

2. Step 2: Match the syntax with blocking assignment

   The statement output = input + 1; uses '=' which is the blocking assignment syntax.

3. Final Answer:

   output = input + 1; -> Option B

4. Quick Check:

   Blocking assignment uses '=' inside always blocks [OK]

Hint: Inside always @(*), use '=' for blocking assignments [OK]

Common Mistakes:

• Using '<=' instead of '=' inside always @(*)
• Using 'assign' inside procedural blocks
• Using 'initial' for combinational logic

3. What is the output of the following Verilog code snippet?

reg a, b, c;
always @(*) begin
  a = 1'b0;
  b = a;
  c = b;
end

medium

A. a=0, b=0, c=0

B. a=0, b=1, c=1

C. a=1, b=0, c=0

D. a=1, b=1, c=1

Solution

1. Step 1: Trace the blocking assignments sequentially

   First, a = 0; sets a to 0 immediately. Then b = a; assigns b the current value of a, which is 0. Finally, c = b; assigns c the current value of b, which is 0.

2. Step 2: Confirm all variables have value 0

   All three variables a, b, and c are assigned 0 in order.

3. Final Answer:

   a=0, b=0, c=0 -> Option A

4. Quick Check:

   Blocking assignments update immediately, so all zero [OK]

Hint: Blocking updates happen immediately in order [OK]

Common Mistakes:

• Assuming b or c keep old values
• Confusing blocking with non-blocking timing
• Ignoring sequential order of assignments

4. Identify the error in this combinational logic block:

always @(*) begin
  out <= in1 & in2;
  out2 = out | in3;
end

medium

A. Using non-blocking (<=) and blocking (=) mixed in combinational block

B. Missing sensitivity list

C. Using continuous assignment inside always block

D. No error, code is correct

Solution

1. Step 1: Check assignment types inside always @(*)

   Combinational logic should use blocking assignments (=) consistently inside always @(*) blocks.

2. Step 2: Identify mixed assignment usage

   The code uses non-blocking (<=) for out and blocking (=) for out2, which is incorrect for combinational logic.

3. Final Answer:

   Using non-blocking (<=) and blocking (=) mixed in combinational block -> Option A

4. Quick Check:

   Combinational blocks use only blocking (=) [OK]

Hint: Use only '=' in combinational always blocks [OK]

Common Mistakes:

• Mixing <= and = inside combinational blocks
• Forgetting to use always @(*) for combinational logic
• Using continuous assign inside procedural blocks

5. You want to design a combinational logic block that calculates the maximum of two inputs a and b. Which of the following always @(*) blocks correctly uses blocking assignments to implement this?

hard

A. always @(*) begin
if (a > b)
max <= a;
else
max <= b;
end

B. always @(a or b) begin
if (a > b)
max = a;
else
max = b;
end

C. always @(a or b) begin
if (a > b)
max <= a;
else
max <= b;
end

D. always @(*) begin
if (a > b)
max = a;
else
max = b;
end

Solution

1. Step 1: Confirm the sensitivity list for combinational logic

   Using always @(*) is preferred for combinational logic to automatically include all inputs.

2. Step 2: Check assignment type for combinational logic

   Blocking assignments (=) must be used inside always @(*) blocks for combinational logic.

3. Step 3: Evaluate each option

   always @(*) begin if (a > b) max = a; else max = b; end uses always @(*) and blocking assignments (=), correctly modeling combinational max logic.

4. Final Answer:

   always @(*) begin if (a > b) max = a; else max = b; end -> Option D

5. Quick Check:

   Combinational max needs always @(*) and '=' [OK]

Hint: Use always @(*) with '=' for combinational max logic [OK]

Common Mistakes:

• Using non-blocking (<=) in combinational always blocks
• Using explicit sensitivity list instead of @(*)
• Mixing blocking and non-blocking assignments