what is blocking assignment in verilog

Blocking vs. Nonblocking in Verilog

The concept of Blocking vs. Nonblocking signal assignments is a unique one to hardware description languages. The main reason to use either Blocking or Nonblocking assignments is to generate either combinational or sequential logic. In software, all assignments work one at a time. So for example in the C code below:

The second line is only allowed to be executed once the first line is complete. Although you probably didn’t know it, this is an example of a blocking assignment. One assignment blocks the next from executing until it is done. In a hardware description language such as Verilog there is logic that can execute concurrently or at the same time as opposed to one-line-at-a-time and there needs to be a way to tell which logic is which.

<=     Nonblocking Assignment

=      Blocking Assignment

The always block in the Verilog code above uses the Nonblocking Assignment, which means that it will take 3 clock cycles for the value 1 to propagate from r_Test_1 to r_Test_3. Now consider this code:

See the difference? In the always block above, the Blocking Assignment is used. In this example, the value 1 will immediately propagate to r_Test_3 . The Blocking assignment immediately takes the value in the right-hand-side and assigns it to the left hand side. Here’s a good rule of thumb for Verilog:

In Verilog, if you want to create sequential logic use a clocked always block with Nonblocking assignments. If you want to create combinational logic use an always block with Blocking assignments. Try not to mix the two in the same always block.

Nonblocking and Blocking Assignments can be mixed in the same always block. However you must be careful when doing this! It’s actually up to the synthesis tools to determine whether a blocking assignment within a clocked always block will infer a Flip-Flop or not. If it is possible that the signal will be read before being assigned, the tools will infer sequential logic. If not, then the tools will generate combinational logic. For this reason it’s best just to separate your combinational and sequential code as much as possible.

One last point: you should also understand the semantics of Verilog. When talking about Blocking and Nonblocking Assignments we are referring to Assignments that are exclusively used in Procedures (always, initial, task, function). You are only allowed to assign the reg data type in procedures. This is different from a Continuous Assignment . Continuous Assignments are everything that’s not a Procedure, and only allow for updating the wire data type.

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Blocking and Non-blocking Assignment in Verilog

• Assignment is only done in procedural block(always ot initial block)
• Both combintational and sequential circuit can be described.
• Assignment can only possible to reg type irrespect of circuit type

Let's say we want to describe a 4-bit shift register in Verilog. For this, we are required to declare a 3-bit reg type variable.

The output of shift[0] is the input of shift[1], output of shift[1] is input of shift[2], and all have the same clock. Let's complete the description using both assignment operator.

Non-Blocking Assignment

When we do synthesis, it consider non-blocking assignment separately for generating a netlist. If we see register assignment in below Verilog code, all register are different if we consider non-blocking assignment separately. If you do the synthesis, it will generate 3 registers with three input/output interconnects with a positive edge clock interconnect for all register. Based on the Verilog description, all are connected sequentially because shift[0] is assigned d, shift[1] is assigned shift[0], and shift[2] is assigned shift[1].

Blocking Assignment

If we use blocking assignment and do the syhtheis, the synthesis tool first generate netlist for first blocking assignment and then go for the next blocking assignment. If in next blocking assignment, if previous output of the register is assigned to next, it will generate only a wire of previously assigned register.

In below Verilog code, even though all looks three different assignment but synthesis tool generate netlist for first blocking assigment which is one register, working on positive edge of clock, input d and output shift[0]. Since blocking assignment is used, for next blocking assignment, only wire is generated which is connected to shift[0]. Same is for next statement a wire is generated which is connected to shift[0].

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Comments (1)

hey in blocking assignment do we get shift in data i dont think so . we get all values same and equal to d.

Please do not focus on the module name; focus on how the netlist is generated after the synthesis.

Procedural Assignments

Blocking assignments, race around condition: a problem with blocking assignment.

Blocking (immediate) and Non-Blocking (deferred) Assignments in Verilog

There are Two types of Procedural Assignments in Verilog.

• Blocking Assignments
• Nonblocking Assignments

To learn more about Delay: Read  Delay in Assignment (#) in Verilog

Blocking assignments

• Blocking assignments (=) are done sequentially in the order the statements are written.
• A second assignment is not started until the preceding one is complete. i.e, it blocks all the further execution before it itself gets executed.

Non-Blocking assignments

• Nonblocking assignments (<=), which follow each other in the code, are started in parallel.
• The right hand side of nonblocking assignments is evaluated starting from the completion of the last blocking assignment or if none, the start of the procedure.
• The transfer to the left hand side is made according to the delays. An intra- assignment delay in a non-blocking statement will not delay the start of any subsequent statement blocking or non-blocking. However normal delays are cumulative and will delay the output.
• Non-blocking schedules the value to be assigned to the variables but the assignment does not take place immediately. First the rest of the block is executed and the assignment is last operation that happens for that instant of time.

To learn more about Blocking and Non_Blocking Assignments: Read Synthesis and Functioning of Blocking and Non-Blocking Assignments

The following example shows  interactions  between blocking  and non-blocking for simulation only (not for synthesis).

For Synthesis (Points to Remember):

• One must not mix “<=” or “=” in the same procedure.
• “<=” best mimics what physical flip-flops do; use it for “always @ (posedge clk..) type procedures.
• “=” best corresponds to what c/c++ code would do; use it for combinational procedures.

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Verification Guide

SystemVerilog Blocking assignment

Blocking assignment.

Blocking assignment statements execute in series order. Blocking assignment blocks the execution of the next statement until the completion of the current assignment execution.

Blocking assignment example

In Below Example, a and b is initialized with value 10 and 15 respectively, after that b is being assigned to a (a value will become 15), and value 20 is assigned to b. After assignment value of a = 15 and b=20.

Simulator Output:

Blocking assignment example-2

In Below Example, a and b are initialized with value 10 and 15 respectively, after that b is being assigned to a (a value will become 15), and value 20 is assigned to b. After assignment value of a = 15 and b = 20.

❮ Previous Next ❯

Verilog: Continuous & Procedural Assignments

Continuous Assignment

Continuous assignment is used to drive a value on to a net in dataflow modeling. The net can be a vector or scalar, indexed part select, constant bit or part select of a vector. Concatenation is also supported with scalar vector types.

module Conti_Assignment (addr1,addr2,wr,din,valid1,valid2,dout); input [31:0] addr1,addr2; input [31:0] din; output [31:0] dout; input valid1,valid2,wr;

wire valid; wire [31:0] addr;

//Net (scalar) continuous assignment assign valid = valid1 | valid2;

//Vector continuous assignment assign addr[31:0] = addr1[31:0] ^ addr2[31:0];

//Part select & Concatenation in Continuous assignment assign dout[31:0] = (valid & wr) ? {din[31:2],2'b11} : 32'd0;

Regular & Implicit Assignment

Regular continuous assignment means, the declaration of a net and its continuous assignments are done in two different statements. But in implicit assignment, continuous assignment can be done on a net when it is declared itself. In the below example, `valid` is declared as wire during the assignment. If signal name is used to the left of the continuous assignment, an implicit net declaration will be inferred. In the below code `dout` is not declared as net, but it is inferred during assignment.

module Implicit_Conti_Assignment (addr1,addr2,wr,din,valid1,valid2,dout); input [31:0] addr1,addr2; input [31:0] din; output [31:0] dout; input valid1,valid2,wr;

//Net (scalar) Implict continuous assignment wire valid = (valid1 | valid2);

//Implicit net declaration -dout assign dout[31:0] = (valid & wr) ? {din[31:2],2'b11} : 32'd0;

Procedural Assignment

We have already seen that continuous assignment updates net, but procedural assignment update values of reg, real, integer or time variable. The constant part select, indexed part select and bit select are possible for vector reg.

There are two types of procedural assignments called blocking and non-blocking. Blocking assignment, as the name says, gets executed in the order statements are specified. The "=" is the symbol used for blocking assignment representation. Non-blocking assignment allows scheduling of assignments. It will not block the execution. The symbol "<=" is used for non-blocking assignment representation and mainly used for concurrent data transfers.

Following example shows the differences in the simulation result by using blocking and non-blocking assignments.

/* module Nonblocking_Assignment (addr1,addr2,wr,din,valid1,valid2,data,aout); input [31:0] addr1,addr2; input [31:0] din; output [31:0] data,aout; input valid1,valid2,wr;

reg [31:0] data,aout, addr; reg valid;

always @(addr1,addr2,wr,din,valid1,valid2) begin valid <= (valid1 | valid2); addr <= (addr1[31:0] | addr2[31:0]); data <= (valid & wr) ? {din[31:2],2'b11} : 32'd0; aout <= wr ? addr: {addr1[15:0],addr2[31:16]}; end initial $monitor($time,"NON-BLOCKING: Values valid1=%b, valid2=%b, wr=%b, addr1=%d, addr2=%d, data=%d, aout=%d", valid1,valid2,wr,addr1,addr2,data,aout); endmodule */ module Blocking_Assignment (addr1,addr2,wr,din,valid1,valid2,data,aout); input [31:0] addr1,addr2; input [31:0] din; output [31:0] data,aout; input valid1,valid2,wr;

always @(addr1,addr2,wr,din,valid1,valid2) begin valid = (valid1 | valid2); addr = (addr1[31:0] | addr2[31:0]); data = (valid & wr) ? {din[31:2],2'b11} : 32'd0; aout = wr ? addr : {addr1[15:0],addr2[31:16]}; $monitor($time,"BLOCKING: Values valid1=%b, valid2=%b, wr=%b, addr1=%d, addr2=%d, data=%d, aout=%d", valid1,valid2,wr,addr1,addr2,data,aout); end endmodule

module test; reg valid1,valid2,wr; reg [31:0] addr1,addr2,din; wire [31:0] data,aout;

Blocking_Assignment Block_Assign(addr1,addr2,wr,din,valid1,valid2,data,aout);

//Nonblocking_Assignment Nonblock_Assign(addr1,addr2,wr,din,valid1,valid2,data,aout);

initial begin valid1 = 0; valid2 = 0; addr1 = 32'd12; addr2 = 32'd36; din = 32'd198; wr = 1;

#5 valid1 = 1; #10 valid1 = 0; valid2 = 1; #10 addr1 = 32'd0; addr2 = 32'd0; #5 wr = 0; #12 wr = 1;

/* ncsim> run 0NON-BLOCKING: Values valid1=0, valid2=0, wr=1, addr1= 12, addr2= 36, data= X, aout= x 5NON-BLOCKING: Values valid1=1, valid2=0, wr=1, addr1= 12, addr2= 36, data= 0, aout= 44 15NON-BLOCKING: Values valid1=0, valid2=1, wr=1, addr1= 12, addr2= 36, data= 199, aout= 44 25NON-BLOCKING: Values valid1=0, valid2=1, wr=1, addr1= 0, addr2= 0, data= 199, aout= 44 30NON-BLOCKING: Values valid1=0, valid2=1, wr=0, addr1= 0, addr2= 0, data= 0, aout= 0 42NON-BLOCKING: Values valid1=0, valid2=1, wr=1, addr1= 0, addr2= 0, data= 199, aout= 0 ncsim: *W,RNQUIE: Simulation is complete. */

/* ncsim> run 0BLOCKING: Values valid1=0, valid2=0, wr=1, addr1= 12, addr2= 36, data= 0, aout= 44 5BLOCKING: Values valid1=1, valid2=0, wr=1, addr1= 12, addr2= 36, data= 199, aout= 44 15BLOCKING: Values valid1=0, valid2=1, wr=1, addr1= 12, addr2= 36, data= 199, aout= 44 25BLOCKING: Values valid1=0, valid2=1, wr=1, addr1= 0, addr2= 0, data= 199, aout= 0 30BLOCKING: Values valid1=0, valid2=1, wr=0, addr1= 0, addr2= 0, data= 0, aout= 0 42BLOCKING: Values valid1=0, valid2=1, wr=1, addr1= 0, addr2= 0, data= 199, aout= 0 ncsim: *W,RNQUIE: Simulation is complete. ncsim> exit */

Verilog Blocking & Non-Blocking assignments elaborated

• Post author By Kevin
• Post date 20 October 2020

Blocking / Non-Blocking assignment rules

The main reason to use either Blocking or Non-Blocking assignments is to generate either combinational or sequential logic.

In non-blocking assignments (<=), all registers inside the always block are updated at the end. In blocking assignments (=), the registers are updated immediately.

Whether or not a flip-flop is inferred from a blocking assignment depends on whether or not the value of the variable being assigned needs to be remembered from one clock edge to the next.

It is good practice to separate combinational and sequential code as much as possible. In verilog, if we want to create sequential logic can use a clocked always block with non-blocking assignments. If on the other hand we want to create combinational logic can use an always block with blocking assignments. Best not to mix the two in the same always block but if they are mixed, need to be careful when doing this. Its up to the synthesis tools to determine whether a blocking assignment within a clocked always block will infer a flip-flop or not. If the signal is read before being assigned (eg fig2 below), the tools will infer sequential logic.

For simplicity purposes only showing in the verilog examples below the Always Block. These Always blocks are blocks of sequential logic since it involves a clock.

If on an active clock edge, the variable tmp is being assigned a value before it’s value is used (ie ‘write before read’ case) then no flip-flop is required & synthesis will not infer it as shown in fig1 below.

If the value of the reg is used before a new value is assigned to it (ie ‘read before write’ case), then the value that is used will be the value that was assigned on a previous clock. Therefore a flip-flop is required here as shown in fig2 below.

If all non-blocking assignments are used within the always block, it will look like :

Non-blocking assignments always imply flip-flops (order of assignments doesn’t matter). Same block diagram is inferred on both cases as shown in fig3 above. They result in simultaneous or parallel statement execution.

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Verilog Assignments

Variable declaration assignment, net declaration assignment, assign deassign, force release.

• Procedural continuous

Legal LHS values

An assignment has two parts - right-hand side (RHS) and left-hand side (LHS) with an equal symbol (=) or a less than-equal symbol (<=) in between.

The RHS can contain any expression that evaluates to a final value while the LHS indicates a net or a variable to which the value in RHS is being assigned.

Procedural Assignment

Procedural assignments occur within procedures such as always , initial , task and functions and are used to place values onto variables. The variable will hold the value until the next assignment to the same variable.

The value will be placed onto the variable when the simulation executes this statement at some point during simulation time. This can be controlled and modified the way we want by the use of control flow statements such as if-else-if , case statement and looping mechanisms.

An initial value can be placed onto a variable at the time of its declaration as shown next. The assignment does not have a duration and holds the value until the next assignment to the same variable happens. Note that variable declaration assignments to an array are not allowed.

If the variable is initialized during declaration and at time 0 in an initial block as shown below, the order of evaluation is not guaranteed, and hence can have either 8'h05 or 8'hee.

Procedural blocks and assignments will be covered in more detail in a later section.

Continuous Assignment

This is used to assign values onto scalar and vector nets and happens whenever there is a change in the RHS. It provides a way to model combinational logic without specifying an interconnection of gates and makes it easier to drive the net with logical expressions.

Whenever b or c changes its value, then the whole expression in RHS will be evaluated and a will be updated with the new value.

This allows us to place a continuous assignment on the same statement that declares the net. Note that because a net can be declared only once, only one declaration assignment is possible for a net.

Procedural Continuous Assignment

• assign ... deassign
• force ... release

This will override all procedural assignments to a variable and is deactivated by using the same signal with deassign . The value of the variable will remain same until the variable gets a new value through a procedural or procedural continuous assignment. The LHS of an assign statement cannot be a bit-select, part-select or an array reference but can be a variable or a concatenation of variables.

These are similar to the assign - deassign statements but can also be applied to nets and variables. The LHS can be a bit-select of a net, part-select of a net, variable or a net but cannot be the reference to an array and bit/part select of a variable. The force statment will override all other assignments made to the variable until it is released using the release keyword.

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Blocking assignments in always block verilog?

now I know in Verilog, to make a sequential logic you would almost always have use the non-blocking assignment (<=) in an always block. But does this rule also apply to internal variables? If blocking assignments were to be used for internal variables in an always block would it make it comb or seq logic?

So, for example, I'm trying to code a sequential prescaler module. It's output will only be a positive pulse of one clk period duration. It'll have a parameter value that will be the prescaler (how many clock cycles to divide the clk) and a counter variable to keep track of it.

I have count's assignments to be blocking assignments but the output, q to be non-blocking. For simulation purposes, the code works; the output of q is just the way I want it to be. If I change the assignments to be non-blocking, the output of q only works correctly for the 1st cycle of the parameter length, and then stays 0 forever for some reason (this might be because of the way its coded but, I can't seem to think of another way to code it). So is the way the code is right now behaving as a combinational or sequential logic? And, is this an acceptable thing to do in the industry? And is this synthesizable?

• 1 Does this answer your question? When does verilog use values from the current and when from the previous timeslot? –  dave_59 Commented Jul 16, 2020 at 6:29
• Synthesize it and see for yourself what it produces. Then ask yourself if that's what you want. –  TomServo Commented Jul 16, 2020 at 21:40

You should follow the industry practice which tells you to use non-blocking assignments for all outputs of the sequential logic. The only exclusion are temporary vars which are used to help in evaluation of complex expressions in sequential logic, provided that they are used only in a single block.

In you case using 'blocking' for the 'counter' will cause mismatch in synthesis behavior. Synthesis will create flops for both q and count . However, in your case with blocking assignment the count will be decremented immediately after it is being assigned the prescaled value, whether after synthesis, it will happen next cycle only.

So, you need a non-blocking. BTW initializing 'count' within declaration might work in fpga synthesis, but does not work in schematic synthesis, so it is better to initialize it differently. Unless I misinterpreted your intent, it should look like the following.

You do not need temp vars there, but you for the illustration it can be done as the following:

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How are blocking statements synthesised? - Verilog

I understand that with the following Verilog code

It uses non-blocking statements all in parallel and I understand that when this is synthesised, it's basically 3 registers in series and it takes 3 clock cycles for 1'b1 to reach r_Test_3.

But what about this,

This uses blocking statements and so, all of this code should be performed in series. How exactly will this be synthesised? I mean will it be the exact same? I'm confused.

• \$\begingroup\$ I would not break my head over it as that code should not be used at all. In a posedge clock block you should only use non-blocking assignments. \$\endgroup\$ –  Oldfart Commented Oct 8, 2018 at 15:37
• \$\begingroup\$ I think you may have confused the issue by assigning 1'b1 to the first variable. I don't think that was your intention. \$\endgroup\$ –  dave_59 Commented Oct 8, 2018 at 16:18

3 Answers 3

What makes VHDL and verilog somewhat confusing is that they were originaly designed to describe hardware for simulation and later re-used to describe hardware for synthisis.

I understand that with the following Verilog code (snip code example using blocking assignments) It uses non-blocking statements all in parallel and I understand that when this is synthesised, it's basically 3 registers in series and it takes 3 clock cycles for 1'b1 to reach r_Test_3.

Remember the initial state of registers is undefined. As a result unless you have specified the initial state of the registers* it is very likely your synthisizer will optimise this to all your signals having a constant value of 1.

Lets for now pretend that you did set the initial state of the registers to zero and you just haven't included that in your example. In that case as you say there will be three registers set one after another on the first three clock edges.

But what about this, (snip code example using blocking assignments) This uses blocking statements and so, all of this code should be performed in series. How exactly will this be synthesised? I mean will it be the exact same? I'm confused.

As photon says that code is equivilent to.

But none of that really answers your real question which boils down to.

How are blocking statements in sequential always blocks synthesised?

How a blocking statement in a sequential always block behaves and hence how it is synthisized depends on where you read it. There are three cases.

• Same always block, read after write.
• Same always block, read before write (or without a write on this invocation of the block).
• Different always block.

In the first case, there is no need for a register (though one may be generated initially and then removed for having zero fanout). The signal just feeds through combinatorially.

In the second case the value needs to be stored from one clock cycle to the next, hence a register is needed.

Note that in some cases flow control may mean that a read is sometimes after a write and sometimes not, in that case the synthisis tool will need to generate a mux to either read from the register or directly from the logic as appropriate, just as it needs to generate muxes when a value is written on multiple different paths.

In the third case the behaviour is not well-defined. In simulation the results will depend on what order the always blocks are evaluated in. In synthisis the tool will probablly spit out a warning and do something, but what it does may not be what you wanted.

Hence you should avoid the third case, if you use blocking assignments in sequential always blocks you should only read the results of those assignments from within the same always block. Signals passing between different always blocks (or going out to the outside world) should always use nonblocking assignments.

* Unfortunately the ability to specify the initial state of registers is something that varies between tools (and versions of tools), some allow use of "initial" blocks for this purpose, some require tool-specific techniques, some don't support it at all.

Your second code block is equivalent to

It might be synthesized as 3 flip-flops, all with inputs tied to logic high.

Or it might just be synthesized as a single flip-flop, with all other logic that is connected to r_Test_1 , r_Test_2 , or r_Test_3 actually being connected to the same physical circuit net (assuming fan-out requirements can still be met this way).

That said, I agree with the comment by Oldfart. Stick to non-blocking assignment for synthesizable sequential logic.

• \$\begingroup\$ Ah alright. So I was right in thinking that it's a bit weird to synthesise that right? \$\endgroup\$ –  AlfroJang80 Commented Oct 8, 2018 at 16:04
• \$\begingroup\$ @AlfroJang80, this example should be easy for a computer to synthesize, but things like this could get confusing for a person to understand how it will synthesize if it gets more complex. Since most Verilog users are more used to seeing non-blocking assignments, it's better to stick to that convention. \$\endgroup\$ –  The Photon Commented Oct 8, 2018 at 16:31
• 1 \$\begingroup\$ The rule to stick to is that the results of a blocking assignment in a sequential always block should only ever be read in that same always block, reading them from other always blocks leads to ambiguity. \$\endgroup\$ –  Peter Green Commented Oct 8, 2018 at 19:30

You first block is equivalent to parallel blocks

Which synthesizes into a chain of flops

You second sysnthesizes to

Which is one flop. This assumes there are no other references to the other variables outside the block.

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Top 25 Verilog Interview Questions You Should Know

• June 12, 2024

Verilog is very important in the VLSI industry, particularly for designing and verifying digital circuits used in modern electronics. Verilog provides a standardized way to describe the functionality and structure of digital circuits. This allows engineers to create a clear and concise representation of the hardware before it’s physically built. Cracking your next Verilog interview can be a breeze with a strong understanding of these essential Verilog interview questions.

With Verilog being very important for the VLSI industry, the skills are highly sought after in the job market, particularly for ASIC/FPGA design and verification engineers. Having Verilog expertise can increase your competitiveness and open doors to new opportunities.

Enrolling in a comprehensive VLSI course can equip one with the knowledge and expertise needed. mastering these top Verilog interview questions will put you ahead of the competition.

The following are a few Verilog questions that can help aspiring engineers to crack their interview questions!

Verilog Interview Questions

1. what is verilog and what is it used for.

Verilog is a hardware description language (HDL) used to describe the digital logic and behavior of electronic circuits. It operates at a register-transfer level (RTL), meaning it focuses on the functionality of the hardware rather than the exact transistor-level implementation. Verilog is widely used in the design and simulation of digital circuits, including microprocessors, FPGAs (Field-Programmable Gate Arrays), and other integrated circuits (ICs).

2. Explain the difference between wire and reg in Verilog.

Both wire and reg are used to declare nets and variables in Verilog, but they have distinct purposes:

• wire: Represents a single physical wire in the circuit. Its value is continuously updated based on the logic driving it. Wires are typically used for combinational logic where the output depends on the current inputs.
• reg: Represents a register that can store a value. Unlike wires, registers hold their value until they are assigned a new one using an always block or an assignment statement. Registers are commonly used for sequential logic where the output depends on both current and past inputs.

3. What are blocking and non-blocking assignments in Verilog?

Verilog assignments can be categorized as blocking or non-blocking based on how they affect the simulation flow:

• Blocking assignment (=): This assignment statement halts the simulation at the point of assignment and calculates the new value before proceeding further. Subsequent statements only execute after the blocking assignment is complete.
• Non-blocking assignment (<=): This assignment schedules the update for the next simulation delta cycle. The simulation continues without interruption, and the new value will be reflected in the next delta cycle.

4. Differentiate between == and === operators in Verilog.

Both == and === are used for comparison in Verilog, but they differ in terms of handling unknown values (represented by x):

• == (equality operator): This operator compares only the bit values (0 or 1) and ignores unknown values (x). It returns 1 if both operands are equal (both 0 or 1) and 0 otherwise. Unknown values (x) can lead to unexpected results with ==.
• === (strict equality operator) : This operator performs a bit-wise comparison and considers unknown values (x). It returns 1 only if all bits are identical (including unknown values) and 0 otherwise. === is generally preferred for reliable comparisons where unknown values might be present.

5. How do you write a Verilog code for a D-latch?

Here’s an example Verilog code for a D-latch:

module DLatch(

  input clk,

  output reg Q

  always @(posedge clk) begin

    Q <= D;

This code defines a module called DLatch with three ports: D (data input), clk (clock signal), and Q (output latch). The always block is triggered on the positive edge of the clock (posedge clk). Inside the block, the current value of the D input is assigned to the register Q using a non-blocking assignment (<=). This ensures that the latch updates its output value on the next clock cycle.

6. Differentiate between Verilog and VHDL?

7. What is a continuous assignment?

A continuous assignment in VHDL defines how a circuit’s output depends on its current inputs. It uses the ‘assign’ keyword and continuously evaluates an expression whenever an input changes. This is ideal for describing combinational logic (e.g., adders) where the output relies directly on the current input combination.

8. Explain the terms $monitor, $display and $strobe.

$monitor, $display, and $strobe are all VHDL system tasks for printing during simulation. Here’s a quick breakdown:

• $monitor: Continuously prints info and signal values whenever a monitored signal changes. Great for observing ongoing behavior.
• $display: Prints a message only once when the statement is encountered. Ideal for specific points like initialization.
• $strobe: Prints a message once at the end of the current simulation time step. Useful for consolidated info after each cycle.

9. What is PLI in Verilog?

PLI (Programming Language Interface) in Verilog acts like a bridge. It lets you call C/C++ functions directly from your Verilog code. This expands Verilog’s abilities for complex tasks like:

• Interfacing with external hardware beyond Verilog’s built-in support.
• Performing advanced calculations more efficiently than native Verilog.
• Creating custom debugging or analysis tools tailored to your needs.

10. What is a sensitivity list?

A sensitivity list in Verilog is like a watchlist for an always block. It tells the simulator which signal changes trigger a re-evaluation of the block’s statements, ensuring the code runs only when necessary.

11. In Verilog, which will be updated first? Variable and Signal?

In Verilog, a signal will always be updated first compared to a variable within the same simulation delta cycle. This behavior stems from the fundamental differences between signals and variables:

Signals: Represent physical wires in the circuit. Assignments using the <= operator schedule the update for the next delta cycle. The actual value change occurs after all concurrent evaluations within that delta cycle are complete.

Variables: Represent temporary storage locations within a process. Assignments using the = operator update the variable’s value immediately.

12. What does timescale 1 Ns/1 Ps mean?

A timescale declaration in Verilog (timescale) defines two key aspects of your simulation:

• Time Unit: The 1ns in this case specifies the basic unit of time used during simulation. This means that any time value you use in your code (delays, event occurrences) will be interpreted relative to nanoseconds (ns).
• Time Precision: The 1ps in this case indicates the smallest time increment that can be represented during simulation. This means that Verilog can distinguish between delays or events that differ by as little as 1 picosecond (ps).

13. Explain the steps involved in writing an FSM code

• Define states and transitions
• Choose representation
• Define Verilog code structure
• Implement state logic
• Implement output logic
• Verification

14. What is transport delay?

In Verilog, transport delay is a modeling concept used to represent the time it takes for a signal to propagate through a wire or gate within a digital circuit. It essentially introduces a latency between the change in an input signal and the corresponding change appearing at the output. Verilog uses the # symbol followed by a time value (e.g., #5ns) within an assignment statement to model transport delay. This delays the assignment of a new value to the target signal by the specified time.

15. What is inertial delay?

Inertial delay is a more advanced concept compared to transport delay for modeling signal propagation. Inertial delay considers the stability of input signals before propagating the change to the output. It ensures the new output value reflects a stable input for a certain duration. Modeling inertial delay can be more complex compared to transport delay.

16. Explain blocking and non-blocking assignments

In Verilog, blocking assignments (=) and non-blocking assignments (<=) are used within always blocks. Blocking assignments evaluate the right-hand side and update the left-hand side immediately, affecting subsequent statements. Non-blocking assignments schedule updates to occur at the end of the current delta cycle, allowing multiple updates to be applied simultaneously. Blocking is used for sequential logic, and non-blocking is used for parallel updates to signals.

17. Explain the concepts of freeze and drive

Freeze and drive are concepts related to forcing signal values during Verilog simulation, but they’re not built-in Verilog commands. Freeze typically refers to forcing a signal to a specific value and keeping it constant throughout the simulation. The value remains frozen until explicitly released. Drive might imply forcing a signal to a specific value but potentially allowing it to change later based on the simulation flow. 

18. Explain the concept of concurrency in Verilog

Verilog is concurrent by nature. This means multiple always blocks, continuous assignments, and procedural blocks can execute seemingly “at the same time” within a simulation delta cycle. The simulator manages the order, ensuring proper evaluation based on dependencies. This allows modeling of parallel hardware behavior efficiently.

19. How do you handle asynchronous resets in Verilog designs?

To handle asynchronous resets in Verilog designs:

• Include the reset signal in the sensitivity list of your always block (@(posedge clk or posedge reset)).
• Assign a low value (usually 0) to the reset signal’s active state.
• Within the always block, use an if statement to prioritize the reset signal. When the reset is active, set the desired initial state for your logic regardless of the clock or other inputs.
• This ensures the reset takes effect immediately, overriding any ongoing operations.

20. Explain the differences between Verilog and SystemVerilog.

Verilog is primarily a Hardware Description Langauge (HDL) for describing the structure and behavior of digital circuits. It works excellent for design implementation. SystemVerilog is a superset of Verilog, offering both HDL and Hardware Verification Language (HVL) capabilities. It expands on Verilog with features for advanced verification including object-oriented programming constructs, and advanced constructs for testbench development. 

21. How can you implement a memory module (e.g., RAM) in Verilog?

Here’s how to implement a simple RAM in Verilog (5 lines):

module RAM #(parameter ADDRESS_WIDTH=4, DATA_WIDTH=8) (

  input we,  // Write enable

  input [ADDRESS_WIDTH-1:0] addr,

  input [DATA_WIDTH-1:0] data_in,

  output reg [DATA_WIDTH-1:0] data_out

  reg [DATA_WIDTH-1:0] mem [2**ADDRESS_WIDTH-1:0];

    if (we) mem[addr] <= data_in;

    data_out <= mem[addr];

22. Describe how to perform verification coverage analysis for a Verilog design.

Verilog verification coverage analysis involves:

• Defining coverage points : Using SystemVerilog covergroups (or similar constructs), you specify signal value combinations or conditions you want to test (e.g., all control signal values, FSM state transitions).
• Collecting coverage data: During simulation, the testbench tracks how often each coverage point is encountered, indicating which design parts have been exercised.
• Analyzing coverage: Coverage reports show uncovered sections (i.e., points not reached), helping identify areas needing more test cases in your testbench.

23. What are generate blocks used for in Verilog?

Generate blocks in Verilog offer two main functionalities:

• Conditional Code Inclusion: Based on parameters or logic, you can include or exclude specific code blocks within the generate statement. This allows creating variations of a module based on configuration settings.
• Loop-based Hardware Replication: Using loops (for or while) within generate blocks, you can efficiently instantiate repetitive hardware structures like memory arrays, adders, or decoders based on a parameter defining the number of elements.

24. In Verilog, what do the casex and casez statements mean?

Verilog’s casex and casez statements are used for bit-wise comparisons within conditional blocks. They allow matching patterns with “don’t care” conditions for unknown or unspecified bits during signal value comparisons.

• casex: Treats X (unknown) as a wildcard that can match any value (0 or 1).
• casez: Treats both X and Z (high-impedance) as wildcards for matching.

25. How does a Verilog loop work?

Verilog loops provide a way to execute a block of code multiple times. This statement is executed only once at the beginning of the loop. It’s typically used to set up a loop counter variable. This statement is executed after each iteration of the loop body. It’s commonly used to increment or decrement the loop counter to control the number of repetitions.

The above Verilog interview questions will be very helpful in preparing for technical Verilog interview questions, as they cover a wide range of essential topics and concepts relevant to digital design and hardware description languages. 

To learn more about Verilog interview questions and other important topics in VLSI take a look at VLSI online courses offered by ChipEdge, an established VLSI training institute in Bangalore .

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