Using $display and $monitor in Verilog - Step-by-Step Execution

Concept Flow - Using $display and $monitor

Start simulation

↓

$display executes once

↓

$monitor sets up continuous watch

↓

Variables change?

↓

$monitor prints new values

↓

Simulation ends

The simulation starts, $display prints once immediately, $monitor watches variables and prints whenever they change until simulation ends.

Execution Sample

Verilog

module test;
  reg a, b;
  initial begin
    $display("Initial values: a=%b, b=%b", a, b);
    $monitor("At time %0t: a=%b, b=%b", $time, a, b);
    a = 0; b = 0;
    #5 a = 1;
    #5 b = 1;
    #5 $finish;
  end
endmodule

This code shows $display printing initial values once, and $monitor printing values whenever a or b changes.

Execution Table

Step	Time	Action	a	b	Output
1	0	$display prints initial values	x	x	Initial values: a=x, b=x
2	0	$monitor starts watching	x	x	At time 0: a=x, b=x
3	0	Assign a=0, b=0	0	0	At time 0: a=0, b=0
4	5	Change a=1	1	0	At time 5: a=1, b=0
5	10	Change b=1	1	1	At time 10: a=1, b=1
6	15	Simulation ends	1	1	No output, simulation finishes

💡 Simulation ends at time 15 with $finish

Variable Tracker

Variable	Start	After Step 3	After Step 4	After Step 5	Final
a	x	0	1	1	1
b	x	0	0	1	1

Key Moments - 3 Insights

Why does $display print only once but $monitor prints multiple times?

$display runs once at the moment it is called (see Step 1), while $monitor sets up a watch and prints every time any watched variable changes (see Steps 3,4,5).

What does the 'x' mean in the initial variable values?

'x' means unknown or uninitialized value in Verilog, shown at simulation start before assignments (Step 1 and 2).

Why does $monitor print at time 0 even before variables are assigned?

$monitor prints immediately when set up with current values (Step 2), even if variables are still unknown.

Visual Quiz - 3 Questions

Look at the execution table, what is the value of 'a' at Step 4?

A 1

B x

C 0

D Unknown

💡 Hint

Check the 'a' column at Step 4 in the execution_table.

At which step does $monitor print the values after 'b' changes?

A Step 3

B Step 5

C Step 4

D Step 6

💡 Hint

Look for when 'b' changes and $monitor output in execution_table.

If we remove $monitor, what happens to the outputs after Step 1?

A $display prints repeatedly

B Outputs continue as normal

C No further outputs after Step 1

D Simulation stops immediately

💡 Hint

Refer to the difference between $display and $monitor in key_moments and execution_table.

Concept Snapshot

$display prints once immediately when called.
$monitor prints whenever watched variables change.
Variables start as 'x' (unknown) until assigned.
$monitor prints initial values at setup.
Use $finish to end simulation.

Full Transcript

This example shows how $display and $monitor work in Verilog. $display prints a message once at the time it is called, showing initial variable values which are unknown (x) at start. $monitor sets up a watch on variables and prints their values whenever they change. Initially, $monitor prints the current values even if unknown. As variables 'a' and 'b' get assigned new values at different times, $monitor prints updated values. The simulation ends with $finish. This helps track variable changes during simulation.

Practice

1. What is the main difference between $display and $monitor in Verilog?

easy

A. $display prints continuously; $monitor prints only once.

B. $display prints once when called; $monitor prints whenever a variable changes.

C. $display is used for debugging; $monitor is used for synthesis.

D. $display prints to a file; $monitor prints to the console.

Solution

1. Step 1: Understand $display behavior

   $display prints a message only once at the time it is called during simulation.

2. Step 2: Understand $monitor behavior

   $monitor automatically prints a message every time any of the monitored variables change value.

3. Final Answer:

   $display prints once when called; $monitor prints whenever a variable changes. -> Option B

4. Quick Check:

   $display = once, $monitor = on change [OK]

Hint: Remember: display = one-time print, monitor = auto on change [OK]

Common Mistakes:

• Thinking $display prints continuously
• Confusing $monitor with file output
• Assuming $monitor prints only once

2. Which of the following is the correct syntax to print the value of a 4-bit register reg [3:0] count; using $display?

easy

A. $display("Count value: %d", count);

B. $display("Count value: %h", count);

C. $display("Count value: %s", count);

D. $display("Count value: %b", count);

Solution

1. Step 1: Identify the data type and format specifier

   The register count is 4-bit binary, so binary format specifier %b is appropriate.

2. Step 2: Match format specifier with $display syntax

   $display("Count value: %b", count); correctly prints the binary value of count.

3. Final Answer:

   $display("Count value: %b", count); -> Option D

4. Quick Check:

   Binary format = %b [OK]

Hint: Use %b for binary, %d decimal, %h hex in $display [OK]

Common Mistakes:

• Using %s which is for strings, not numbers
• Using %d for binary values
• Forgetting to include variable after format string

3. Given the following Verilog code snippet, what will be the output during simulation?

reg [1:0] a;
initial begin
  a = 2'b00;
  $monitor("Time=%0t a=%b", $time, a);
  #5 a = 2'b01;
  #5 a = 2'b11;
  #5 a = 2'b10;
end

medium

A. Prints four lines showing time and value of 'a' at 0, 5, 10, and 15 time units.

B. Prints only one line at time 0 with initial value of 'a'.

C. Prints three lines at times 5, 10, and 15 but not at time 0.

D. No output because $monitor is not called correctly.

Solution

1. Step 1: Understand $monitor initial print

   $monitor prints immediately when called (time 0) and then whenever monitored variables change.

2. Step 2: Trace changes to 'a' and timing

   'a' changes at 5, 10, and 15 time units, so $monitor prints at these times plus initial time 0.

3. Final Answer:

   Prints four lines showing time and value of 'a' at 0, 5, 10, and 15 time units. -> Option A

4. Quick Check:

   $monitor prints on call + changes = 4 prints [OK]

Hint: $monitor prints immediately and on every change [OK]

Common Mistakes:

• Assuming $monitor prints only on changes, missing initial print
• Thinking $monitor prints only once
• Confusing $display behavior with $monitor

4. Identify the error in this Verilog code snippet:

reg clk;
initial begin
  $monitor("Clock = %b", clk);
  clk = 0;
  #10 clk = 1;
  #10 clk = 0;
end

medium

A. clk should be declared as wire, not reg, for $monitor to work.

B. Missing semicolon after $monitor statement causes syntax error.

C. $monitor is called before clk is initialized, so initial print shows unknown value.

D. No error; code runs and prints correctly.

Solution

1. Step 1: Check order of $monitor and variable initialization

   $monitor is called before clk is assigned a value, so initial print shows 'x' (unknown).

2. Step 2: Understand $monitor initial print behavior

   $monitor prints immediately on call, so it prints clk's value before initialization.

3. Final Answer:

   $monitor is called before clk is initialized, so initial print shows unknown value. -> Option C

4. Quick Check:

   Initialize variables before $monitor call [OK]

Hint: Initialize variables before $monitor to avoid unknown prints [OK]

Common Mistakes:

• Assuming $monitor waits for variable initialization
• Thinking missing semicolon causes error here
• Confusing reg and wire for $monitor usage

5. You want to monitor two signals reg [3:0] data; and reg valid; so that the simulation prints their values only when valid changes. Which is the best way to do this using $monitor?

hard

A. $monitor("Data=%b Valid=%b", data, valid);

B. Use an if statement inside $monitor to print only when valid changes.

C. Use $monitor with only valid variable: $monitor("Valid=%b", valid); and add $display for data.

D. Use $monitor with both variables but assign data only when valid changes.

Solution

1. Step 1: Understand $monitor behavior with multiple variables

   $monitor prints whenever any monitored variable changes. Monitoring both data and valid prints on either change.

2. Step 2: Restrict printing to valid changes only

   To print only when valid changes, monitor only valid. But then data won't print updated values automatically.

3. Step 3: Best practical approach

   Monitoring both variables prints all changes, but since data changes only when valid changes, this works well. $monitor("Data=%b Valid=%b", data, valid); is simplest and correct.

4. Final Answer:

   $monitor("Data=%b Valid=%b", data, valid); -> Option A

5. Quick Check:

   Monitor all needed vars; prints on any change [OK]

Hint: Monitor all variables to track changes together [OK]

Common Mistakes:

• Trying to put conditions inside $monitor (not allowed)
• Monitoring only one variable and missing others' changes
• Using $display inside $monitor incorrectly