When to use blocking (combinational) in Verilog - Step-by-Step Execution

Concept Flow - When to use blocking (combinational)

Start Combinational Logic

↓

Use Blocking Assignments (=)

↓

Evaluate statements in order

↓

Update variables immediately

↓

Output combinational signals

↓

End Combinational Logic

Blocking assignments (=) are used in combinational logic to evaluate statements in order and update variables immediately, reflecting real combinational behavior.

Execution Sample

Verilog

always @(*) begin
  a = b + c;
  d = a + 1;
end

This code uses blocking assignments to compute 'a' then 'd' in order, modeling combinational logic.

Execution Table

Step	Statement	Variable Before	Evaluation	Variable After
1	a = b + c;	b=2, c=3, a=0	a = 2 + 3	a=5
2	d = a + 1;	a=5, d=0	d = 5 + 1	d=6
3	End always block	a=5, d=6	No change	a=5, d=6

💡 All statements executed in order with blocking assignments, variables updated immediately.

Variable Tracker

Variable	Start	After Step 1	After Step 2	Final
a	0	5	5	5
d	0	0	6	6
b	2	2	2	2
c	3	3	3	3

Key Moments - 2 Insights

Why do we use blocking assignments (=) in combinational logic?

Because blocking assignments update variables immediately in order, matching how combinational logic works, as shown in execution_table steps 1 and 2.

What happens if we use non-blocking assignments (<=) in combinational logic?

Non-blocking assignments delay updates until the end of the time step, which can cause incorrect intermediate values in combinational logic, unlike the immediate updates seen in the execution_table.

Visual Quiz - 3 Questions

Look at the execution_table, what is the value of 'a' after step 1?

A 0

B 5

C 6

D 2

💡 Hint

Refer to the 'Variable After' column in row for step 1.

At which step does 'd' get its final value?

A Step 2

B Step 1

C Step 3

D Never

💡 Hint

Check the 'Variable After' column for 'd' in each step.

If we replaced '=' with '<=' in the code, how would the variable updates change?

A Variables update immediately in order

B Variables update randomly

C Variables update only at the end of the block

D Variables do not update

💡 Hint

See key_moments explanation about non-blocking assignments delaying updates.

Concept Snapshot

Use blocking assignments (=) in combinational always blocks.
Statements execute in order, variables update immediately.
This models real combinational logic behavior.
Avoid non-blocking (<=) in combinational logic to prevent incorrect delays.
Example:
always @(*) begin a = b + c; d = a + 1; end

Full Transcript

In Verilog, blocking assignments (=) are used in combinational logic blocks to ensure statements execute in order and variables update immediately. This matches how combinational circuits work in hardware. For example, in the code 'always @(*) begin a = b + c; d = a + 1; end', 'a' is updated first, then 'd' uses the new value of 'a'. The execution table shows step-by-step how variables change. Using non-blocking assignments (<=) here would delay updates and cause incorrect behavior. Remember to use blocking assignments for combinational logic to model immediate updates and ordered execution.

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