Blocking vs non-blocking assignment in Verilog - Visual Side-by-Side Comparison

Concept Flow - Blocking vs non-blocking assignment

Start

↓

Execute blocking (=) assignment

↓

Variable updated immediately

↓

Execute next statement

↓

Execute non-blocking (<=) assignment

↓

Schedule update for end of time step

↓

Execute next statement

↓

At end of time step, update variable

↓

End

Blocking assignments update variables immediately and pause execution; non-blocking assignments schedule updates to happen later, allowing parallel updates.

Execution Sample

Verilog

reg a, b, c;
always @(*) begin
  a = b;      // blocking
  c <= a;     // non-blocking
end

Shows blocking assignment updating 'a' immediately, then non-blocking assignment scheduling 'c' to update after the time step.

Execution Table

Step	Statement	Variable Updated	Value Before	Value After	Notes
1	a = b;	a	0	1	Blocking assignment updates 'a' immediately to 'b' value (1)
2	c <= a;	c	0	0 (scheduled)	Non-blocking schedules 'c' to update later with current 'a' (1)
3	End of time step	c	0	1	'c' updated to 'a' value at end of time step

💡 All statements executed; non-blocking assignments update at end of time step

Variable Tracker

Variable	Start	After Step 1	After Step 2	After Step 3 (End)
a	0	1	1	1
b	1	1	1	1
c	0	0	0 (scheduled)	1

Key Moments - 2 Insights

Why does 'a' change immediately but 'c' does not after their assignments?

Because 'a' uses blocking assignment (=), it updates immediately as shown in step 1 of execution_table. 'c' uses non-blocking (<=), so its update is scheduled and only happens at the end of the time step (step 3).

What happens if we use blocking assignments for both 'a' and 'c'?

Both variables update immediately in order, so 'c' would get the new value of 'a' right away, unlike the non-blocking case where 'c' updates later.

Visual Quiz - 3 Questions

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

A Undefined

B 0

C 1

D Scheduled update

💡 Hint

Check the 'Value After' column for 'a' in step 1 of execution_table.

At which step does 'c' actually get its new value?

A Step 3

B Step 1

C Step 2

D Never

💡 Hint

Look at the 'Notes' column for 'c' in execution_table; non-blocking updates happen at end of time step.

If both assignments were blocking, what would happen to 'c' after step 2?

A 'c' would remain unchanged

B 'c' would update immediately to 'a's new value

C 'c' would update at end of time step

D 'c' would get 'b's value directly

💡 Hint

Blocking assignments update variables immediately, so 'c' would get 'a's updated value right away.

Concept Snapshot

Blocking assignment (=): updates variable immediately, execution waits.
Non-blocking assignment (<=): schedules update at end of time step, allows parallel updates.
Use blocking for combinational logic, non-blocking for sequential logic.
Order matters for blocking; non-blocking updates happen together after all statements.

Full Transcript

This visual trace shows how blocking and non-blocking assignments work in Verilog. Blocking assignments update variables immediately and pause execution until done. Non-blocking assignments schedule updates to happen later, at the end of the current time step, allowing multiple updates to happen in parallel. In the example, 'a' is assigned from 'b' using blocking, so 'a' changes right away. 'c' is assigned from 'a' using non-blocking, so 'c' keeps its old value during the statements and updates only after all statements finish. This difference is important for writing correct hardware logic.